I am looking for an example of a function $f:\mathbb R^2\to\mathbb R$ such that there is a point $x\in\mathbb R^2$ with the following properties:
1) All partial derivatives of second order exist in a neighborhood of $x$.
2) At least one of those partial derivatives is not continuous in $x$.
3) The Hessian matrix of $f$ in $x$ is symmetric.
I think it should be possible to find such a function but I wasn't very successful in finding one. If we drop the first property it is easy, with it however, I didn't find any example yet. Any help appreciated.
$$ f(x,y) = \begin{cases} 0 & \text{if }x=y=0 \\ \bigl(\frac{2xy}{x^2+y^2}\bigr)^2 & \text{otherwise} \end{cases} $$
This is obviously smooth in $\mathbb R^2\setminus\{(0,0)\}$, and both first-order partial derivatives are 0 everywhere on the axes. So the Hessian at the origin is zero.
Since $f(t,t)=1$ but $f(0,t)=0$ for all $t\ne 0$, $\frac{\partial f}{\partial x}$ must grow unboundedly large close to $(0,0)$, and is therefore not continuous there.
(Note that for nonzero $x$ and $y$ we have $f(x,y)=\sin^2(2\arctan(y/x))$.)