$$\left(\frac{\partial P}{\partial T}\right)_V=-\left(\frac{\partial V}{\partial T}\right)_P\left(\frac{\partial P}{\partial V}\right)_T.$$

It comes from the way partial derivatives are handled through Jacobi determinants. (Remember, the simple chain rule is for ordinary derivatives.)

On the left-hand side, you express $P$ in terms of $T$ and $V$. There is the identity,

$$\frac{\partial(P,V)}{\partial(T,V)}= \left(\frac{\partial P}{\partial T}\right)_V \left(\frac{\partial V}{\partial V}\right)_T - \left(\frac{\partial P}{\partial V}\right)_T \left(\frac{\partial V}{\partial T}\right)_V=\left(\frac{\partial P}{\partial T}\right)_V,$$

since

$$\left(\frac{\partial x}{\partial x}\right)_y=1,\qquad{\rm and}\qquad\left(\frac{\partial y}{\partial x}\right)_y=0.$$

On the right-hand side, you are expressing everything in terms of $P$ and $T$, which amounts to transforming the above identity as follows:

\begin{align} \left(\frac{\partial P}{\partial T}\right)_V&=\frac{\partial(P,V)}{\partial(T,V)}=\frac{\partial(P,V)}{\partial(P,T)}\frac{\partial(P,T)}{\partial(T,V)}\\ &=\left[\left(\frac{\partial P}{\partial P}\right)_T\left(\frac{\partial V}{\partial T}\right)_P-\left(\frac{\partial P}{\partial T}\right)_P\left(\frac{\partial V}{\partial P}\right)_T\right] \left[\left(\frac{\partial P}{\partial T}\right)_V\left(\frac{\partial T}{\partial V}\right)_T-\left(\frac{\partial P}{\partial V}\right)_T\left(\frac{\partial T}{\partial T}\right)_V\right]\\ &=-\left(\frac{\partial V}{\partial T}\right)_P\left(\frac{\partial P}{\partial V}\right)_T. \end{align}