a function with differentiable partial derivatives but unequal mixed derivatives I am looking for an example of a function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ such that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are both differentiable at some point, say the origin, but $\frac{\partial^2 f}{\partial x\partial y}\neq\frac{\partial^2 f}{\partial y\partial x}$ at that point.
 A: If $f_{xy}$ and $f_{yx}$ are continuous, then they are necessarily equal.
The only examples you will find where the mixed partial derivatives are $not$ equal will be just like the standard example, $f(x,y)=x \cdot y\cdot \displaystyle \frac{x^2-y^2}{x^2+y^2}$ (and $f(0,0)=0$), and this example works precisely because $f_{xy}$ and $f_{yx}$ are not continuous at the origin. In this case $f_{xy}(0,0)=1$ and $f_{yx}(0,0)=-1$. At all other points, the mixed partials are equal.
A: The original "standard" counterexample most certainly has continuous first partials everywhere.
The function is proportional to $r^2 \sin(4 \theta)$, which — since $\sin$ is bounded — is dominated by $r^2$ at the origin. Hence all directional derivatives are $0$ there, and approach $0$ along any curve into the origin.  (And from the rational function expression for this, it's clear that the function is infinitely differentiable everywhere else.)
A: There is no such example!
Theorem: If a function is twice-differentiable (meaning that its first partial derivatives are differentiable), then its mixed second partial derivatives are symmetric.
Proof: see https://ncatlab.org/nlab/show/differentiable+map#symmetry.
So symmetry of mixed second partial derivatives does not require the second partial derivatives to be continuous; the first partial derivatives just have to be differentiable.  In other words, the function doesn't have to be twice continuously differentiable, just twice differentiable.
Corollary: If a function's second partial derivatives are continuous, then its mixed second partial derivatives are symmetric.
Proof: Apply the theorem that if a function's partial derivatives are continuous, then the function is differentiable, to the first partial derivatives of the original function; then apply the preceding theorem.
So the famous theorem about symmetry of partial derivatives is just a corollary of this more basic theorem.  (Although there's also a version of this theorem that only requires one of the mixed second partial derivatives to be continuous, and that one's not just a corollary.)
A: Let $z = \arctan(y/x)$, then $\frac{d}{dy}\left(\frac{dz}{dx}\right) \not= \frac{d}{dx}\left( \frac{dz}{dy}\right)$
