Bivariate density function For a distribution function $F$ density function is defined as $\dfrac{\partial^2 F}{\partial x \,\partial y}$. Is it essential that $F$ is differentiable?
Is it required that $\dfrac{\partial^2 F}{\partial x \, \partial y}=\dfrac{\partial^2 F}{\partial y \, \partial x}$?
 A: The density is not defined that way; rather a theorem states that that is what it is, provided certain reasonable assumptions hold.
With a univariate c.d.f. $G$, it is not always true that the density is $G'$.  In particular, that is false when $G$ is the c.d.f. of a discrete distribution, and then $G'$ would not be differentiable --- indeed it is not continuous.  That the density is $G'$ fails even in some cases where $G$ is continuous: google the term "Cantor distribution".
I've never thought about the question of whether there can be a c.d.f. so badly behaved that its mixed partial derivatives are not equal.  But if hypotheses hold that entail that the density must be equal to each of those two second derivatives, then necessarily they must be equal to each other because both are equal to the density.  A question arises of whether just one of them might be equal to the density.  That one I'd have to work on for a while.
A: If you allow for discrete or mixed discrete/continuous probability densities, then 
$\partial^2F \over {\partial x \partial y}$ need not exist at points (or curves) of finite probability.  For example, with domain of the unit square $0<x<1, 0<y<1$, let's look at 
$$F(x,y) = \frac{2}{3} \left( 1 + \theta(y-x) \right) $$
where $\theta(u)$ is the step function, with value $1$ for $u>0$ and zero otherwise. The mixed second derivative is non-existant on $y=x$ yet this is a perfectly good cumulative distribution function.
As to  $\frac{\partial^2 F}{\partial x \, \partial y}=\frac{\partial^2 F}{\partial y \, \partial x}$, that is not a requirement per se, but the same pathology that causes this unnatural inequality might well also disqualify the function as a cumulative distribution function.
