Folland Advanced Calculus Ex. 2.5.7 
Suppose that the variables $E$, $T$, $V$, and $P$ are related by a pair of equations, $f(E,T,V,P)=0$  and $g(E,T,V,P)=0$, that can be solved for any two of the variables in terms of the other two, and suppose that the differential equation $\partial_VE-T\partial_TP+P=0$ is satisfied when $V$ and $T$ are taken as the independent variables. Show that $\partial_PE+T\partial_TV+P\partial_PV=0$ when $P$ and $T$ are taken as the independent variables.

If we put $\phi(V,T)=(E(V,T),T,V,P(V,T))$ and $F=(f,g)$, then $(F\circ\phi)(V,T)=0$ and hence $F'(\phi(V,T))\phi'(V,T)=0$.
How to proceed any further?

Taking $V$  and $T$  as independent variables means that $E$ and $P$
are determined as functions of $V$ and $T$.
Let $E=\eta(V, T) $ and $P=\zeta(V,T)$
Then given $\partial_VE-T\partial_TP+P=0$ becomes $\partial_1\eta -T\partial_2\zeta+P=0\quad \ldots (1) $
Let us take $P, T$ as independent variables.

Then differentiating $E=\eta(V, T) $ and $P=\zeta(V,T)$ w.r.to $P$ yields$$\partial_PE =(\partial_1\eta )(\partial_PV)$$
and $$ 1=(\partial_1\zeta )(\partial_PV) $$
Hence $$\partial_1\eta =\frac{\partial_PE}{\partial_PV}$$
and $$\partial_1\zeta =\frac{1}{\partial_PV}$$

Again differentiating  $P=\zeta(V,T)$ w.r.to $T$ yields $$0 =(\partial_1\zeta )(\partial_TV)+\partial_2\zeta$$
Hence $\partial_2\zeta=-(\partial_1\zeta )(\partial_TV)=-\frac{\partial_TV}{\partial_PV}$
Now substituting the values of $\partial_1\eta, \partial_2\zeta$ in $(1) $ , we have
$\begin{align}0&=\frac{\partial_PE}{\partial_PV}-T ( -\frac{\partial_TV}{\partial_PV})+P\\&=\partial_PE +T \partial_TV+P\partial_TV\end{align}$
