Ideal gas derivations
This looks like the differential of the ideal gas equation. I'm going to use $V_m$ for the molar volume.
$$PV_m=RT$$
Now we convert each side to the differential. We keep $P$ and $V_m$ together since they are both variable. We can move $R$ out of the differential because it is a constant.
$$d\left(PV_m\right)=RdT$$
We can evaluate $d\left(PV_m\right)$ using the product rule:
$$d\left(PV_m\right) = PdV_m + V_mdP$$
Now:
$$PdV_m + V_mdP = RdT$$
Which is equivalent to your first equation.
**However, at constant temperature $dT=0$, so at constant temperature, the equation simplifies to:
$$PdV_m + V_mdP=0$$
$$V_mdP = - PdV_m$$
$$\frac{dP}{dV_m}=-\frac{P}{V_m}$$
Which is not equal to your second equation. The closest I can get is the following:
$$\frac{dP}{dV_m}=-\frac{P}{V_m}\cdot\frac{V_m}{V_m}=-\frac{PV_m}{V^2_m}=-\frac{RT}{V^2_m}$$
Which is what we would get by evaulating:
$$\left(\frac{\partial}{\partial V_m}\right)_T \left(P=\frac{RT}{V_m}\right)$$