Answer
$$
\left(P+\frac{n^{2} a}{V^{2}}\right)(V-n b)=n R T
$$
$\Rightarrow$
$$
\frac{\partial T}{\partial P}=\frac{V-n b}{n R},
$$
and
$$
\begin{aligned}
\frac{\partial P}{\partial V}=\frac{2 n^{2} a}{V^{3}}-\frac{n R T}{(V-n b)^{2}}
\end{aligned}
$$
Work Step by Step
$$
\left(P+\frac{n^{2} a}{V^{2}}\right)(V-n b)=n R T \quad\quad\quad\quad (1)
$$
$\Rightarrow$
$$
T=\frac{1}{n R}\left(P+\frac{n^{2} a}{V^{2}}\right)(V-n b),
$$
To find $\frac{\partial T}{\partial P}$, we differentiate implicitly with respect to $P$, being careful to treat $V$ as a constant:
$$
\frac{\partial T}{\partial P}=\frac{1}{n R}(1)(V-n b)=\frac{V-n b}{n R}.
$$
Now, we rewrite Eq. (1) as follows
$$
P+\frac{n^{2} a}{V^{2}}=\frac{n R T}{V-n b}
$$
$\Rightarrow$
$$
P=\frac{n R T}{V-n b}-\frac{n^{2} a}{V^{2}}=n R T(V-n b)^{-1}-n^{2} a V^{-2}.
$$
To find $\frac{\partial P}{\partial V}$, we differentiate explicitly with respect to $V$, being careful to treat $T$ as a constant:
$$
\begin{aligned}
\frac{\partial P}{\partial V}&=-n R T(V-n b)^{-2}(1)+2 n^{2} a V^{-3} \\
&=\frac{2 n^{2} a}{V^{3}}-\frac{n R T}{(V-n b)^{2}}.
\end{aligned}
$$
