Find a differentiable function $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ such that $f_{xy}\neq f_{yx}$. Find a differentiable function $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ such that the second-order partial derivatives $f_{xy}(0, 0)$ and $f_{yx}(0, 0)$ exist but are not equal.
So I need a function of $x$ and $y$ where $f_{xy}(0, 0)\ne f_{yx}(0, 0)$. 
So looking at $f(x,y)=x^2y^2$:
$f_x(x,y)=2xy^2$
$f_{xy}(x,y)=4xy$
$f_y(x,y)=2x^2y$
$f_{yx}(x,y)=4xy$
This function obviously fails, but I'm trying to find an example and I'm having trouble. I obviously need an some terms added, but the added terms will be killed off by the second partial derivative... So any hints?
 A: As explained in the comments, because of the Schwarz's theorem, the function cannot have continuous second partial derivatives at $(0,0)$.
Here there is an example.
And here (page 20) there is a reference for the following interesting facts.


*

*There exists a function $f$, the mixed second derivatives of which exist at every point but such that $f_{xy}\neq f_{yx}$ on a set of positive measure.

*There exists a function $f$, the mixed second derivatives of which exist almost everywhere and such that $f_{xy}\neq f_{yx}$ almost everywhere.
A: Let $~~f(x,y)=
\begin{cases}
\frac{xy\left(x^2-y^2\right)}{x^2+y^2},~ & \text{if}\ (x,y) \ne  (0,0),\\
0, & \text{if}\  \text{if}\ (x,y) =  (0,0). \\
\end{cases}\\
$
We know by definition that
$\[f_{xy}(a,b)=\frac{\partial}{\partial x}\left(f_y(a,b)\right)=\lim_{h\to 0} \frac{f_y(a+h,b)-f_y(a,b)}{h}.\]$
\begin{equation}\label{pd.cdtwo16}
\text{Therefore, we have }~~~~~~~~~~f_{xy}(0,0)=\lim_{h\to 0} \frac{f_y(h,0)-f_y(0,0)}{h}~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation}
We have
[f_{y}(h,0)=\lim_{k\to 0} \frac{f(h,k)-f(h,0)}{k}=\lim_{k\to 0} \frac{ \frac{ hk\left(h^2-k^2\right)}{h^2+k^2}-0} {k}]
[=\lim_{k\to 0} \frac{ h\left(h^2-k^2\right)}{h^2+k^2} =\frac{h^3}{h^2}=h.]
Also [f_{y}(0,0)=\lim_{k\to 0} \frac{f(0,k)-f(0,0)}{k}=\lim_{k\to 0} \frac{ 0-0}{k}=0.]
Therefore, \eqref{pd.cdtwo16} becomes
[f_{xy}(0,0)=\lim_{h\to 0} \frac{h-0}{h}=1.]
Now, we have
[f_{yx}(a,b)=\frac{\partial}{\partial y}\left(f_x(a,b)\right)=\lim_{k\to 0} \frac{f_x(a,b+k)-f_x(a,b)}{k}.]
\begin{equation}\label{pd.cdtwo17}
\text{Therefore, we have }~~~~~~~~~~f_{yx}(0,0)=\lim_{k\to 0} \frac{f_x(0,k)-f_x(0,0)}{k}~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation}
We have
[f_{x}(0,k)=\lim_{h\to 0} \frac{f(h,k)-f(0,k)}{h}=\lim_{h\to 0} \frac{ \frac{ hk\left(h^2-k^2\right)}{h^2+k^2}-0} {k}]
[=\lim_{h\to 0} \frac{ k\left(h^2-k^2\right)}{h^2+k^2} =\frac{-k^3}{k^2}=-k.]
Also [f_{x}(0,0)=\lim_{h\to 0} \frac{f(h,0)-f(0,0)}{h}=\lim_{h\to 0} \frac{ 0-0}{h}=0.]
Therefore, \eqref{pd.cdtwo17} becomes
[f_{yx}(0,0)=\lim_{k\to 0} \frac{0-k}{k}=-1.]
Thus, we have shown that $f_{xy}(0,0) \ne f_{yx}(0,0).\diamond$
