$P(n,k)$ is the number of partitions of $n$ into exactly $k$ parts. (It is also the number of partitions of $n$ into any number of parts with greatest part equal to $k$; also, for $n\ge k$, the number of partitions of $n-k$ into parts of size at most $k$.)
$P(n,3)$ is OEIS sequence A069905.
You are probably familiar with the identity
$$P(n,k)=P(n-1,k-1)+P(n-k,k)\quad\quad(n\ge k\ge1);\tag1$$
if not, observe that $P(n-1,k-1)$ is the number of partitions of $n$ into $k$ parts with least part equal to $1$, and $P(n-k,k)$ is the number of partitions of $n$ into $k$ parts with least part greater than $1$. We need this for $k=3$:
$$P(n,3)=P(n-1,2)+P(n-3,3)\quad\quad(n\ge3)\tag2.$$
Since $P(n,2)=\left\lfloor\frac n2\right\rfloor$, we can rewrite $(2)$ as
$$P(n,3)=\left\lfloor\frac{n-1}2\right\rfloor+P(n-3,3)\quad\quad(n\ge3).\tag3$$
Applying $(3)$ twice, we get
$$P(n,3)=\left\lfloor\frac{n-1}2\right\rfloor+\left\lfloor\frac{n-4}2\right\rfloor+P(n-6,3)\quad\quad(n\ge6).\tag4$$
Since
$$\left\lfloor\frac{n-1}2\right\rfloor+\left\lfloor\frac{n-4}2\right\rfloor=\begin{cases}\frac{n-2}2+\frac{n-4}2=n-3\quad\text{ if }n\text{ is even },\\
\frac{n-1}2+\frac{n-5}2=n-3\quad\text{ if }n\text{ is odd },
\end{cases}$$
we can rewrite $(4)$ as
$$P(n,3)=P(n-6,3)+n-3\quad\quad(n\ge6).\tag5$$
Hence, if we suppose that $n\ge6$ and
$$P(n-6,3)=\left\lfloor\frac{(n-6)^2}{12}\right\rfloor=\left\lfloor\frac{n^2-12n+36}{12}\right\rfloor=\left\lfloor\frac{n^2}{12}\right\rfloor-n+3,$$
it follows by $(5)$ that
$$P(n,3)=P(n-6,3)+n-3=\left\lfloor\frac{n^2}{12}\right\rfloor.$$
Since the equality $P(n,3)=\left\lfloor\frac{n^2}{12}\right\rfloor$ holds for the base cases $n=0,1,2,4,5$, it follows by induction that it holds whenever $n\not\equiv3\pmod6$.
Remark. In the same way, we can show that the identity
$$P(n,3)=\left\lfloor\frac{n^2+3}{12}\right\rfloor$$
holds for all $n$ without exception.