Why is the temperature totally differentiable? Suppose there is a function $T=T(p,V)$, called temperature, which is invertible in the whole space according to the pressure $p$ and the volume $V$.
Is it true that $T$ is a twice differentiable function of $p$ and $V$, and that furthermore $\frac{{{\partial }^{2}}T}{\partial p\partial V}=\frac{{{\partial }^{2}}T}{\partial V\partial p}$?
If yes, how can we prove it? If no, what additional criteria are necessary for such condition to hold?
 A: That constraint on the second derivatives with respect to $p$ and $V$ is a direct consequence of the temperature $T$ being a state function of $p$ and $V$.
Physically this means that the values of $p$ and $V$ is all the information you need to compute the temperature for that state of the system.
This may seem an obvious condition to impose but it is not always verified for physical systems.
In fact, in many realistic scenarios increasing first the pression by an amount $\Delta p$ and then the volume by an amount $\Delta V$ you may obtain a different state (with a different temperature $T$), than the state you would have obtained by varying first the volume and then the pressure by the same amounts.
If on the other hand you want to describe a system for which the above is not true, that is, a system for which no matter the way you arrived to the pressure $p$ and volume $V$ you always have the same temperature, then this will impose a condition on the functional dependence of $T$ on $p$ and $V$, which assumes the form you mentioned:
$$ \frac{\partial^2 T}{\partial p\partial V} = \frac{\partial^2 T}{\partial V\partial p}. $$
There are many ways to see why $T$ being a state function implies the above.
Mathematically you can see it as an instance of Schwarz's theorem.
A more intuitive way to see it is to note that
$ \frac{\partial^2 T}{\partial V\partial p}(p,V) $
gives the amount of variation in $T$ if starting from $(p,V)$ you vary $p$ by a small amount $dp$ and then you vary $V$ by a small amount $dV$.
$ \frac{\partial^2 T}{\partial p\partial V}(p,V) $ is the same when varying first $V$ and then $p$, and if you want to impose that it does not matter the order with which you vary the parameters it immediately follows that these two quantities must be the same. 
