Starting from $$dU=TdS-PdV$$ We can write, for instance $U(T,V)$ and $S(T,V)$ to obtain:
$$\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_T dV=T\left(\frac{\partial S}{\partial T}\right)_V dT+T\left(\frac{\partial S}{\partial T}\right)_T dV-PdV$$ Identifying the terms in front of the differential forms we arrive at:
$$\left(\frac{\partial U}{\partial T}\right)_VdT=T\left(\frac{\partial S}{\partial T}\right)_V$$ $$\left(\frac{\partial U}{\partial V}\right)_T =T\left(\frac{\partial S}{\partial T}\right)_T-P$$
This is the usual way I would go about deriving such equalities. What is about to follow is very unformal. I noticed that these expressions can be obtained much faster by dividing by, for instance $dT_V$, where the subscript indicates that $V$ is held constant. With this notation we have $\left(\frac{dU}{dT_V}\right)= \left(\frac{\partial U}{\partial T}\right)_V $
Then, we obtain the following immediately from the first equation:
$$\left(\frac{\partial U}{\partial T}\right)_V=T\left(\frac{\partial S}{\partial T}\right)_V-P\underbrace{\left(\frac{\partial V}{\partial T}\right)_V}_{=0}$$ and for the second equation we divide by $dV_T$: $$\left(\frac{\partial U}{\partial V}\right)_T =T\left(\frac{\partial S}{\partial T}\right)_T-P\underbrace{\left(\frac{\partial V}{\partial V}\right)_T}_{=1}$$ and we obtain the same results with this informal procedure.
My questions are: Does this work in general? Can the same be done for inexact differentials like $\delta Q$? Does this still work when the number of particles is varied, for instance when $U(N,V,P)$?
