Multivariable calculus and Thermodynamics Somewhere in my thermodynamics text I read the following.
$\left(\frac{dU}{dP}\right)_{V} = \left(\frac{\partial U}{\partial T}\right)_{V} \left(\frac{\partial T}{\partial P}\right)_{V}$
Is it correct?
It should be something like this.
$\left(\frac{dU}{dP}\right)_{V} = \left(\frac{\partial U}{\partial T}\right)_{everything\ else} \left(\frac{d T}{d P}\right)_{V}$
$U$ is not usually a function of T or P, since $U \equiv U(S,V,N)$

PS: I have edited the $\left(\frac{\partial U}{\partial T}\right)_{everything}$ to $\left(\frac{\partial U}{\partial T}\right)_{everything\ else}$
 A: Firstly I'd like to confirm L. White's statement that U is a function of P and T. In thermodynamics we are usually dealing with a space that can be described by 3 coordinates (degrees of freedom). You may choose S, V, N since U can be explicitly written in these coordinates. But you may as well replace one of the variables if you know the equation of state of your system, e.g. the ideal gas equation relates V, N, P and T, so you can e.g. replace V.
Secondly about the actual question: there are different ways to think about partial derivatives and total derivatives and different conventions are used in the literature. There is one notation that I find particularly confusing, namely 
$\left(\frac{dU}{dP}\right)_{V}$
If you hold $V$ constant the total derivative really is a partial derivative, since partial derivatives are nothing else than total derivatives with something held constant. Partial derivatives are like taking a derivative along a certain direction in a multi-dimensional vector space. See also this answer of mine. You may correctly note that only holding $V$ constant is not sufficient to define a direction in the vector space. In most undergraduate thermodynamics holding $N$ constant is implied by this notation, which then does define the direction completely. I would suspect that is also what is meant here, but of course I can not be sure since the OP did not state the context.
Then written as a partial derivative the relation in doubt should become clear. And this also shows that the notation $\left(\frac{\partial U}{\partial T}\right)_{everything}$ does not really make sense. If you hold "everything else" constant you can't take a derivative, because the system will be fixed by your equation of state, whatever it may be (e.g. ideal gas law).
A: I am not a mathematician so I am going out on a shaky limb here - I answer in order to possibly learn something - so please saw gently!
In the form it is posed isn't the left side of your first equation mathematically equivalent to the right side in the same manner that 
(3/7) = (3/5) * (5/7)
If not please explain exactly why not so I can understand something from long long ago  - for me at least.
A: U is always and very much a function of T, so your statement that it is not is in error.
from http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/inteng.html
"Internal energy involves energy on the microscopic scale. For an ideal monoatomic gas, this is just the translational kinetic energy of the linear motion of the "hard sphere" type atoms, and the behavior of the system is well described by kinetic theory. However, for polyatomic gases there is rotational and vibrational kinetic energy as well."
So, for monatomic gasses the case is clear - internal energy is equal to translational kinetic energy. Temperature is a measure of the average kinetic energy of translation for the atom, so the internal energy is always a function of temperature.
For polyatomic molecules it is also clear. According to the equipartition theory, the energy stored in the various available modes will be equally divided between those modes and the translational kinetic energy of the molecule, so the energy stored in each mode will be directly related to the kinetic energy of translation. Temperature is a measure of the average kinetic energy of translation for the molecule, so the internal energy is once again always a function of temperature. 
I think correcting this misunderstanding should remove your concern.
