The following is Problem 6.1 from a book I'm self-studying, the "Mathematics of Classical and Quantum Physics", by Byron and Fuller, 1e. Given
$$ \begin{equation} f(z)= \begin{cases} \displaystyle\frac{xy^2(x+iy)}{x^2+y^2}\equiv u+iv,&z\ne0\\ 0&z=0 \end{cases} \end{equation} $$
where $z=x+iy$, the reader is asked to determine where $f(z)$ is analytic and differentiable. My thought process is as follows: according to the book, a function is differentiable at a point where the Cauchy-Riemann conditions are satisfied and the partial derivatives of $u,v$ exist and are continuous. I calculate the partial derivatives to be
$$ \begin{align*} \frac{\partial u}{\partial x}&=\frac{2xy^4}{(x^2+y^2)^2}\\[0.3em] \frac{\partial u}{\partial y}&=\frac{2x^4y}{(x^2+y^2)^2}\\[0.3em] \frac{\partial v}{\partial x}&=\frac{x^2y^3+y^5-2xy^4}{(x^2+y^2)^2}\\[0.3em] \frac{\partial v}{\partial y}&=\frac{3x^3y^2+xy^4}{(x^2+y^2)^2} \end{align*} $$
from which it follows that the Cauchy-Riemann conditions are satisfied at every point along the real axis $y=0$ [alternatively, writing
$$ f(z)=-\frac18\left(\frac{z^3}{z^*}-z^2-z^*z+z^{*}{}^2\right) $$
the choice $z=x\in\mathbb{R}$ makes $f(z)$ independent of $z^*$]. Graphing the partial derivatives suggests they continuously approach zero independently of the way $x,y$ approach the origin. Therefore, since the Cauchy-Riemann conditions are satisfied at every point along the real axis, and the partial derivatives of $u,v$ exist and are continuous there, the function $f(z)$ is differentiable at every point along the real axis. However, it is nowhere analytic, since it is not differentiable in a two-dimensional strip containing the real axis.
Is this line of reasoning correct? Thank you for your time.
