Calculate directional derivative at $(0, 0)$

Question

I need to calculate the partial derivative for the function $f$, that is $f(x, y) = \frac{x^2y}{x^2+y^2}$ when $(x, y) \neq (0,0)$ and $0$ when $(x, y) = (0, 0)$.

I can calculate the partial derivative for $(x, y) \neq (0, 0)$. However, how do I calculate the partial derivative at $(0, 0)$ in the direction of the vector $v = <1, 1>$?

Answer 1

We have

$$ \frac{f(t,t)-f(0,0)}{t}=1$$

for all $t \ne 0.$ Hence

$$\frac{\partial f}{\partial v} (0,0)=1.$$