$$f(x,y) = \left\{\begin{array}{l l} \frac{2xy}{x^2+y^2} & \quad \text{if $(x,y)\neq(0,0),$}\\ 0 & \quad \text{if $(x,y)=(0,0).$} \end{array} \right.$$
I know that this function is a standard example of a discontinuous function whose first order partial derivatives exist. I was just wondering if $\dfrac{\partial^2f}{\partial x \partial y}$ exist at the origin.
My work so far: $$\frac{\partial^2f}{\partial x \partial y}(0,0)=\frac{\partial}{\partial x}\frac{\partial f}{\partial y}(0,0)=\lim_{h\to0} \frac{\frac{\partial f}{\partial y}(h,0)-\frac{\partial f}{\partial y}(0,0)}{h}$$
Sub in $\dfrac{\partial f}{\partial y}=\dfrac{2x(x^2-y^2)}{(x^2+y^2)^2}$ for $(x,y)\neq (0,0)$ and $\dfrac{\partial f}{\partial y}(0,0)=0$, I get $$\frac{\partial^2f}{\partial x \partial y}(0,0)= \lim_{h\to 0} \frac{2}{h^2}=0$$
I don't know whether what I've got is correct. Any help appreciated.