Directional derivative of piecewise defined function?

Question

Let $$f(x,y)=\begin{cases}\frac{x^3+y^3}{x^2-y^2},\ x^2-y^2\neq 0 \\ \ \ \ \ 0 \ \ \ \ ,x^2-y^2=0\end{cases}$$

Then find the directional derivative of $f$ at $(0,0)$ in the direction of vector $\langle\frac45,\frac35 \rangle$.

I don't know how to calculate the directional derivative of piecewise defined function.

How can I solve this ?

Answer 1

You can just use the definition $$ \partial_{(4/5,3/5)}f(0,0) = \lim_{t \to 0} \dfrac{f(0+4t/5, 0 + 3t/5)-f(0,0)}{t} = \frac{13}{5} $$

Answer 2

Let $v=\langle v_1,v_2\rangle=\langle\frac45,\frac35 \rangle$. Then $v_1^2 \ne v_2^2$, hence the directional derivative is given by

$$ \lim_{t \to 0} \frac{f(tv)-f(0,0)}{t}.$$

Can you proceed ?