Q:

Find partial derivative at point $(0,0)$ of $(x^2-y^2)/(x^2+y^2)$

It's a bit wierd question but I have to ask it.
$$ \text{Let }\space f(x, y) =
\begin{cases}
\dfrac{x^2-y^2}{x^2 + y^2},  & \text{if $(x, y) \ne (0,0)$} \\
0, & \text{if $(x, y) = (0, 0)$}  \\
\end{cases}$$
The following question is:

Calculate the partial derivatives of f at the points where they exist.

A:

Obviously the partial derivatives exist at $(x,y) \not = (0,0)$, as it's a rational function and the denominator is never zero. Now to find the partial derivatives at $(0,0)$ you have:
$$\lim_{x \to 0}\frac{f(x,0) - f(0,0)}{x} = \lim_{x \to 0} \frac{\frac{x^2 - 0^2}{x^2 + y^2} - 0}{x} = \lim_{x \to 0} \frac 1x = \infty $$
$$\lim_{y \to 0}\frac{f(0,y) - f(0,0)}{y} = \lim_{y \to 0} \frac{\frac{0^2 - y^2}{0^2 + y^2} - 0}{x} = \lim_{y \to 0} -\frac 1y = -\infty $$
Therefore the function has partial derivatives at every single point except the origin.

