How to evaluate this partial derivative? If we have some function:

$$ f(x,y)= \begin{cases} \frac{xy(x^2-y^2)}{x^2+y^2} &amp;\text {if }(x,y) \neq (0,0) \\
0 &amp; \text{otherwise}\end{cases}$$

then find $$\frac{d^2f}{dxdy}(0,0).$$

This is my question I need to solve. Now I was thinking that it would be $0$ since $f(x,y)=0$ at $(0,0)$ but according to the workings it is not, now I have no idea how to evaluate this derivative and would like some help?

Also the question is posed asking for this  $$\frac{d^2f}{dxdy}(0,0)$$.
 but shouldn't it be a partial derivative or something? I don't know I'm really confused. Any help?
In order to find $\frac{\partial ^2f}{\partial x\partial y}(0,0)$ you first need to find $\frac{\partial f}{\partial y}(t,0)$ for $t$ in a neighborhood of $0$.

$\frac{\partial f}{\partial y}(0,0)$ is easily seen to be $0$ because $f(0,y)=0$ no matter whether $y$ is zero or nonzero.

For nonzero $t$ you only need to look at the first case of the definition of $f$, because that case governs $f(t,y)$ for all possible $y$. Holding $t$ constant, you would use the quotient rule to differentiate with respect to $y$, but since the derivative of the denominator is $0$ one of the terms vanish and we get just
$$ \frac{\partial}{\partial y}\frac{ty(t^2-y^2)}{t^2+y^2} \bigg|_{y=0} 
= \frac{\frac{\partial}{\partial y}ty(t^2-y^2)}{t^2+y^2} \bigg|_{y=0} $$
When the dust clears you get a very simple expression for this which agrees with what we just found for $t=0$ and which is trivial to differentiate with respect to $t$.
Use the quotient rule and differentiate partially for x (treating y as a constant) and then partially for y (treating x as a constant)