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?
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.
Let $~~f(x,y)=
\begin{cases}
\frac{xy\left(x^2-y^2\right)}{x^2+y^2},~ &amp; \text{if}\ (x,y) \ne  (0,0),\\
0, &amp; \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 $f_{xy}(0,0) \ne f_{yx}(0,0).\diamond$