$f(x,y) = 0$ when $x = y = 0$. Shouldn't $\frac{\partial^2 f}{\partial y \partial x} = 0$? I'm given the following function:
$f(x,y) =$
$
\begin{cases}
2xy \frac{x^2 - y^2}{x^2 + y^2} & x^2 + y^2 \neq 0 \\
0 & x=y=0
\end{cases}
$
And the task is to prove that $f_{xy}(0,0) = -2$ and that $f_{yx}(0,0) = 2$.
At first glance I'd say that neither of these is true since at $(0,0)$ the function is 0 and so both expressions should equal 0. I thought maybe I was wrong and tried to work with the other piece of the function but I got the following:
$$
f_{xy}=\frac{\partial^2 f}{\partial y \partial x} = \frac{2(x^6+9x^4y^2-9x^2y^4-y^6)}{(x^2 + y^2)^3}
$$
Which is clearly not $-2$ if we evaluate for $x=0, y=0$.
What am I missing? Am I going about it all wrong? Any help is appreciated.

Let $f$ be given by 
$$f(x,y)=\begin{cases}
2xy\frac{x^2-y^2}{x^2+y^2}&,x^2+y^2>0\\\\
0&,x^2+y^2=0
\end{cases}$$
For $x^2+y^2>0$, we have
$$\begin{align}\frac{\partial f(x,y)}{\partial x}&=\frac{2y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2}\\\\\frac{\partial f(x,y)}{\partial y}&=-\frac{2x(y^4+4x^2y^2-x^4)}{(x^2+y^2)^2}\end{align}$$ 

To calculate the first partial derivaties at the origin, we revert to using the limit definition of the partial derivatives.  

Proceeding, we find that 
$$f_x(0,0)=\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h}=0\\\\
f_y(0,0)=\lim_{h\to 0}\frac{f(0,h)-f(0,0)}{h}=0$$

We proceed in a similar fashion to find the mixed partial derivatives at the origin.  We find that 
$$\begin{align}
f_{yx}(0,0)&=\lim_{h\to 0}\frac{f_y(h,0)-f_y(0,0)}{h}\\\\
&=\lim_{h\to 0}\frac{-\frac{2h(0^4+4x^20^2-h^4)}{(h^2+0^2)^2}-0}{h}\\\\
&=2
\end{align}$$
and
$$\begin{align}
f_{xy}(0,0)&=\lim_{h\to 0}\frac{f_x(0,h)-f_x(0,0)}{h}\\\\
&=\lim_{h\to 0}\frac{\frac{2h(0^4+4x^20^2-h^4)}{(h^2+0^2)^2}-0}{h}\\\\
&=-2
\end{align}$$
as was to be shown!
