I have this piecewise function:
$$f(x,y)= \left\{ \begin{array}{lcc} \frac{(x+y)^2}{x^2+y^2}; & (x,y)\neq(0,0) \\ \\ 1; & (x,y)=(0,0) \\ \\ \end{array} \right.$$
We know that it is not continuous at $(0,0)$ and we are asked about whether this statement about partial derivative is true: $$f_x(0,0)=0$$ I checked that $f$ function and got that it is not differentiable at $(0,0)$. The derivative using the definition based on limits, also shows that this statement is wrong. But my professor wrote that it is true and quals to zero as said. Here is my problem. Do I miss something, or that partial derivative by x really equals to zero?
For $x \neq 0$ $$\frac{f(x,0) - f(0,0)}{x}= \frac{x^2/x^2 -1}{x} = 0.$$
So indeed $f_x(0,0) = 0$.