I was asked to find the partial derivatives at every point for the following function:
$$ f(x,y)= \left\{ \begin{array}{c} \frac{(x^2+y)^2}{x^4+y^2} \qquad (x,y)\neq(0,0) \\ \;\;\; 1 \qquad \quad (x,y)=(0,0) \\ \end{array} \right. $$
Now, for all $(x,y)\neq(0,0)$ (after some legwork) the answers are: $$\frac{\partial f(x,y)}{\partial x}=\frac{4yx(y^2-x^4)}{(x^4+y^2)^2}$$ $$\frac{\partial f(x,y)}{\partial y}=\frac{2x^2(x^4-y^2)}{(x^4+y^2)^2}$$
My question is about when $(x,y)=(0,0)$. Is $f$ discontinuous there? What can I say about the partial derivatives at that point? They don't exist there? They exist but are not differentiable there?
Thank you.