\underline{Summary}Summary
Update
TO answer comment below, Here is the hand solution. The question was just misunderstanding on my part reading result of DSolve, since $\lamda=0$ can not be an eigenvalue, but it also showed there in the solution to DSolve, so I was asking why.
But DSolve is allowed to return trivial solution $y(x)=0$ which is not allowed when finding the eigenvalues. Lesson of the day: Use DEigenvalues to find eigenvalues instead of DSolve.
NDEigenvalues[{-y''[x]+NeumannValue[-y[x],x==1]+NeumannValue[y[x],x==0]},y[x],{x,0,1},6]
L0=1;
tab1=Table[ n^2 Pi^2/L0,{n,{1,2,3,4,5}}]//N;
Join[{-1},tab1]
Hand solution
Assume the solution is $y=Ae^{rx}$, then the characteristic equation is
\begin{align*} r^{2}+\lambda & =0\\ r & =\pm\sqrt{-\lambda} \end{align*}
Assuming $\lambda<0$
In this case $-\lambda$ is positive and hence $\sqrt{-\lambda}$ is also positive. Let $\sqrt{-\lambda}=\mu$ where $\mu>0$. Hence the roots are $\pm \mu$. This gives the solution
$$ y=c_{1}\cosh\left( \mu x\right) +c_{2}\sinh\left( \mu x\right) $$
Hence
$$ y^{\prime}=\mu c_{1}\sinh\left( \mu x\right) +\mu c_{2}\cosh\left( \mu x\right) $$
Left B.C. gives
\begin{equation} 0=c_{1}+\mu c_{2}\tag{1} \end{equation}
Right B.C. gives
\begin{align*} 0 & =c_{1}\cosh\left( \mu L\right) +c_{2}\sinh\left( \mu L\right) +\mu c_{1}\sinh\left( \mu L\right) +\mu c_{2}\cosh\left( \mu L\right) \\ & =\cosh\left( \mu L\right) \left( c_{1}+\mu c_{2}\right) +\sinh\left( \mu L\right) \left( c_{2}+\mu c_{1}\right) \end{align*}
Using (1) in the above, it simplifies to
$$ 0=\sinh\left( \mu L\right) \left( c_{2}+\mu c_{1}\right) $$
But from (1) again, we see that $c_{1}=-\mu c_{2}$ and the above becomes
\begin{align*} 0 & =\sinh\left( \mu L\right) \left( c_{2}-\mu\left( \mu c_{2}\right) \right) \\ & =\sinh\left( \mu L\right) \left( c_{2}-\mu^{2}c_{2}\right) \\ & =c_{2}\sinh\left( \mu L\right) \left( 1-\mu^{2}\right) \end{align*}
But $\sinh\left( \mu^{2}L\right) \neq0$ since $\mu^{2}L\neq0$ and so either $c_{2}=0$ or $\left( 1-\mu^{2}\right) =0$. $c_{2}=0$ results in trivial solution, therefore $\left( 1-\mu^{2}\right) =0$ or $\mu^{2}=1$ but $\mu ^{2}=-\lambda$, hence $\lambda=-1$ is the eigenvalue.
Corresponding eigenfunction is
$$ y=c_{1}\cosh\left( x\right) +c_{2}\sinh\left( x\right) $$
Using (1) the above simplifies to
\begin{align*} y & =-\mu c_{2}\cosh\left( x\right) +c_{2}\sinh\left( x\right) \\ & =c_{2}\left( -\mu\cosh\left( x\right) +\sinh\left( x\right) \right) \end{align*}
But $\mu=\sqrt{-\lambda}=1$, hence the eigenfunction is
$$ \fbox{$y\left( x\right) =c_2\left( -\cosh\left( x\right) +\sinh\left( x\right) \right) $} $$
Assuming $\lambda=0$
Solution now is
$$ y=c_{1}x+c_{2} $$
Therefore
$$ y^{\prime}=c_{1} $$
Left B.C. $0=y\left( 0\right) +y^{\prime}\left( 0\right) $ gives
\begin{equation} 0=c_{2}+c_{1} \tag{2} \end{equation}
Right B.C. $0=y\left( L\right) +y^{\prime}\left( L\right) $ gives
\begin{align*} 0 & =\left( c_{1}L+c_{2}\right) +c_{1}\\ 0 & =c_{1}\left( 1+L\right) +c_{2} \end{align*}
But from (2) $c_{1}=-c_{2}$ and the above becomes
\begin{align*} 0 & =-c_{2}\left( 1+L\right) +c_{2}\\ 0 & =-c_{2}L \end{align*}
Which means $c_{2}=0$ and therefore the trivial solution. Therefore $\lambda=0$ is not an eigenvalue.
Assuming $\lambda>0$
Solution is \begin{equation} y=c_{1}\cos\left( \sqrt{\lambda}x\right) +c_{2}\sin\left( \sqrt{\lambda }x\right) \tag{A} \end{equation}
Hence
$$ y^{\prime}=-\sqrt{\lambda}c_{1}\sin\left( \sqrt{\lambda}x\right) +\sqrt{\lambda}c_{2}\cos\left( \sqrt{\lambda}x\right) $$
Left B.C. gives
\begin{equation} 0=c_{1}+\sqrt{\lambda}c_{2} \tag{3} \end{equation}
Right B.C. gives
\begin{align*} 0 & =c_{1}\cos\left( \sqrt{\lambda}L\right) +c_{2}\sin\left( \sqrt{\lambda}L\right) -\sqrt{\lambda}c_{1}\sin\left( \sqrt{\lambda }L\right) +\sqrt{\lambda}c_{2}\cos\left( \sqrt{\lambda}L\right) \\ & =\cos\left( \sqrt{\lambda}L\right) \left( c_{1}+\sqrt{\lambda} c_{2}\right) +\sin\left( \sqrt{\lambda}L\right) \left( c_{2}-\sqrt {\lambda}c_{1}\right) \end{align*}
Using (3) in the above, it simplifies to
$$ 0=\sin\left( \sqrt{\lambda}L\right) \left( c_{2}-\sqrt{\lambda} c_{1}\right) $$
But from (3), we see that $c_{1}=-\sqrt{\lambda}c_{2}$. Therefore the above becomes \begin{align*} 0 & =\sin\left( \sqrt{\lambda}L\right) \left( c_{2}-\sqrt{\lambda}\left( -\sqrt{\lambda}c_{2}\right) \right) \\ & =\sin\left( \sqrt{\lambda}L\right) \left( c_{2}+\lambda c_{2}\right) \\ & =c_{2}\sin\left( \sqrt{\lambda}L\right) \left( 1+\lambda\right) \end{align*}
Only choice for non trivial solution is either $\left( 1+\lambda\right) =0$ or $\sin\left( \sqrt{\lambda}L\right) =0$. But $\left( 1+\lambda\right) =0$ implies $\lambda=-1$ but we said that $\lambda>0$. Hence other choice is \begin{align*} \sin\left( \sqrt{\lambda}L\right) & =0\\ \sqrt{\lambda}L & =n\pi\qquad n=1,2,3,\cdots\\ \lambda_{n} & =\left( \frac{n\pi}{L}\right) ^{2}\qquad n=1,2,3,\cdots \end{align*}
The above are the eigenvalues. The corresponding eigenfunction is from (A)
$$ y=c_{1_{n}}\cos\left( \sqrt{\lambda_{n}}x\right) +c_{2_{n}}\sin\left( \sqrt{\lambda_{n}}x\right) $$
But $c_{1_{n}}=-\sqrt{\lambda_{n}}c_{2_{n}}$ and the above becomes
\begin{align*} y\left( x\right) & =-\sqrt{\lambda_{n}}c_{2_{n}}\cos\left( \sqrt {\lambda_{n}}x\right) +c_{2}\sin\left( \sqrt{\lambda_{n}}x\right) \\ & =C_{n}\left( -\sqrt{\lambda_{n}}\cos\left( \sqrt{\lambda_{n}}x\right) +\sin\left( \sqrt{\lambda_{n}}x\right) \right) \end{align*}
\underline{Summary}
Eigenvalue $\lambda=-1$ with eigenfunction $y\left( x\right) =c_{2}\left( -\cosh\left( x\right) +\sinh\left( x\right) \right) $ and eigenvalues $\lambda_{n}=\left( \frac{n\pi}{L}\right) ^{2},n=1,2,3,\cdots$ with eigenfunctions $C_{n}\left( -\sqrt{\lambda_{n}}\cos\left( \sqrt{\lambda_{n} }x\right) +\sin\left( \sqrt{\lambda_{n}}x\right) \right) $. Listing the eigenvalues in order:
$$ \lambda=\left\{ -1,\frac{\pi^{2}}{L^{2}},\frac{4\pi^{2}}{L^{2}},\frac {9\pi^{2}}{L^{2}},\frac{16\pi^{2}}{L^{2}},\cdots\right\} $$
Update
TO answer comment below, Here is the hand solution. The question was just misunderstanding on my part reading result of DSolve, since $\lamda=0$ can not be an eigenvalue, but it also showed there in the solution to DSolve, so I was asking why.
But DSolve is allowed to return trivial solution $y(x)=0$ which is not allowed when finding the eigenvalues. Lesson of the day: Use DEigenvalues to find eigenvalues instead of DSolve.
NDEigenvalues[{-y''[x]+NeumannValue[-y[x],x==1]+NeumannValue[y[x],x==0]},y[x],{x,0,1},6]
L0=1;
tab1=Table[ n^2 Pi^2/L0,{n,{1,2,3,4,5}}]//N;
Join[{-1},tab1]
Hand solution
Assume the solution is $y=Ae^{rx}$, then the characteristic equation is
\begin{align*} r^{2}+\lambda & =0\\ r & =\pm\sqrt{-\lambda} \end{align*}
Assuming $\lambda<0$
In this case $-\lambda$ is positive and hence $\sqrt{-\lambda}$ is also positive. Let $\sqrt{-\lambda}=\mu$ where $\mu>0$. Hence the roots are $\pm \mu$. This gives the solution
$$ y=c_{1}\cosh\left( \mu x\right) +c_{2}\sinh\left( \mu x\right) $$
Hence
$$ y^{\prime}=\mu c_{1}\sinh\left( \mu x\right) +\mu c_{2}\cosh\left( \mu x\right) $$
Left B.C. gives
\begin{equation} 0=c_{1}+\mu c_{2}\tag{1} \end{equation}
Right B.C. gives
\begin{align*} 0 & =c_{1}\cosh\left( \mu L\right) +c_{2}\sinh\left( \mu L\right) +\mu c_{1}\sinh\left( \mu L\right) +\mu c_{2}\cosh\left( \mu L\right) \\ & =\cosh\left( \mu L\right) \left( c_{1}+\mu c_{2}\right) +\sinh\left( \mu L\right) \left( c_{2}+\mu c_{1}\right) \end{align*}
Using (1) in the above, it simplifies to
$$ 0=\sinh\left( \mu L\right) \left( c_{2}+\mu c_{1}\right) $$
But from (1) again, we see that $c_{1}=-\mu c_{2}$ and the above becomes
\begin{align*} 0 & =\sinh\left( \mu L\right) \left( c_{2}-\mu\left( \mu c_{2}\right) \right) \\ & =\sinh\left( \mu L\right) \left( c_{2}-\mu^{2}c_{2}\right) \\ & =c_{2}\sinh\left( \mu L\right) \left( 1-\mu^{2}\right) \end{align*}
But $\sinh\left( \mu^{2}L\right) \neq0$ since $\mu^{2}L\neq0$ and so either $c_{2}=0$ or $\left( 1-\mu^{2}\right) =0$. $c_{2}=0$ results in trivial solution, therefore $\left( 1-\mu^{2}\right) =0$ or $\mu^{2}=1$ but $\mu ^{2}=-\lambda$, hence $\lambda=-1$ is the eigenvalue.
Corresponding eigenfunction is
$$ y=c_{1}\cosh\left( x\right) +c_{2}\sinh\left( x\right) $$
Using (1) the above simplifies to
\begin{align*} y & =-\mu c_{2}\cosh\left( x\right) +c_{2}\sinh\left( x\right) \\ & =c_{2}\left( -\mu\cosh\left( x\right) +\sinh\left( x\right) \right) \end{align*}
But $\mu=\sqrt{-\lambda}=1$, hence the eigenfunction is
$$ \fbox{$y\left( x\right) =c_2\left( -\cosh\left( x\right) +\sinh\left( x\right) \right) $} $$
Assuming $\lambda=0$
Solution now is
$$ y=c_{1}x+c_{2} $$
Therefore
$$ y^{\prime}=c_{1} $$
Left B.C. $0=y\left( 0\right) +y^{\prime}\left( 0\right) $ gives
\begin{equation} 0=c_{2}+c_{1} \tag{2} \end{equation}
Right B.C. $0=y\left( L\right) +y^{\prime}\left( L\right) $ gives
\begin{align*} 0 & =\left( c_{1}L+c_{2}\right) +c_{1}\\ 0 & =c_{1}\left( 1+L\right) +c_{2} \end{align*}
But from (2) $c_{1}=-c_{2}$ and the above becomes
\begin{align*} 0 & =-c_{2}\left( 1+L\right) +c_{2}\\ 0 & =-c_{2}L \end{align*}
Which means $c_{2}=0$ and therefore the trivial solution. Therefore $\lambda=0$ is not an eigenvalue.
Assuming $\lambda>0$
Solution is \begin{equation} y=c_{1}\cos\left( \sqrt{\lambda}x\right) +c_{2}\sin\left( \sqrt{\lambda }x\right) \tag{A} \end{equation}
Hence
$$ y^{\prime}=-\sqrt{\lambda}c_{1}\sin\left( \sqrt{\lambda}x\right) +\sqrt{\lambda}c_{2}\cos\left( \sqrt{\lambda}x\right) $$
Left B.C. gives
\begin{equation} 0=c_{1}+\sqrt{\lambda}c_{2} \tag{3} \end{equation}
Right B.C. gives
\begin{align*} 0 & =c_{1}\cos\left( \sqrt{\lambda}L\right) +c_{2}\sin\left( \sqrt{\lambda}L\right) -\sqrt{\lambda}c_{1}\sin\left( \sqrt{\lambda }L\right) +\sqrt{\lambda}c_{2}\cos\left( \sqrt{\lambda}L\right) \\ & =\cos\left( \sqrt{\lambda}L\right) \left( c_{1}+\sqrt{\lambda} c_{2}\right) +\sin\left( \sqrt{\lambda}L\right) \left( c_{2}-\sqrt {\lambda}c_{1}\right) \end{align*}
Using (3) in the above, it simplifies to
$$ 0=\sin\left( \sqrt{\lambda}L\right) \left( c_{2}-\sqrt{\lambda} c_{1}\right) $$
But from (3), we see that $c_{1}=-\sqrt{\lambda}c_{2}$. Therefore the above becomes \begin{align*} 0 & =\sin\left( \sqrt{\lambda}L\right) \left( c_{2}-\sqrt{\lambda}\left( -\sqrt{\lambda}c_{2}\right) \right) \\ & =\sin\left( \sqrt{\lambda}L\right) \left( c_{2}+\lambda c_{2}\right) \\ & =c_{2}\sin\left( \sqrt{\lambda}L\right) \left( 1+\lambda\right) \end{align*}
Only choice for non trivial solution is either $\left( 1+\lambda\right) =0$ or $\sin\left( \sqrt{\lambda}L\right) =0$. But $\left( 1+\lambda\right) =0$ implies $\lambda=-1$ but we said that $\lambda>0$. Hence other choice is \begin{align*} \sin\left( \sqrt{\lambda}L\right) & =0\\ \sqrt{\lambda}L & =n\pi\qquad n=1,2,3,\cdots\\ \lambda_{n} & =\left( \frac{n\pi}{L}\right) ^{2}\qquad n=1,2,3,\cdots \end{align*}
The above are the eigenvalues. The corresponding eigenfunction is from (A)
$$ y=c_{1_{n}}\cos\left( \sqrt{\lambda_{n}}x\right) +c_{2_{n}}\sin\left( \sqrt{\lambda_{n}}x\right) $$
But $c_{1_{n}}=-\sqrt{\lambda_{n}}c_{2_{n}}$ and the above becomes
\begin{align*} y\left( x\right) & =-\sqrt{\lambda_{n}}c_{2_{n}}\cos\left( \sqrt {\lambda_{n}}x\right) +c_{2}\sin\left( \sqrt{\lambda_{n}}x\right) \\ & =C_{n}\left( -\sqrt{\lambda_{n}}\cos\left( \sqrt{\lambda_{n}}x\right) +\sin\left( \sqrt{\lambda_{n}}x\right) \right) \end{align*}
\underline{Summary}
Eigenvalue $\lambda=-1$ with eigenfunction $y\left( x\right) =c_{2}\left( -\cosh\left( x\right) +\sinh\left( x\right) \right) $ and eigenvalues $\lambda_{n}=\left( \frac{n\pi}{L}\right) ^{2},n=1,2,3,\cdots$ with eigenfunctions $C_{n}\left( -\sqrt{\lambda_{n}}\cos\left( \sqrt{\lambda_{n} }x\right) +\sin\left( \sqrt{\lambda_{n}}x\right) \right) $. Listing the eigenvalues in order:
$$ \lambda=\left\{ -1,\frac{\pi^{2}}{L^{2}},\frac{4\pi^{2}}{L^{2}},\frac {9\pi^{2}}{L^{2}},\frac{16\pi^{2}}{L^{2}},\cdots\right\} $$