Problem
Let $f(x,y)= \frac{xy^3}{x^3+y^6}$ if $(x,y)\neq (0,0)$ ,and define $f(0,0)=0$. Prove that the derivative $f'(O,a)$ exists for every vector $a$ and compute its value in terms of the components of $a$.
Attempt
$f'(O,a)= \lim_{x\to 0}\frac{f(ha_1,ha_2)}{h}$ =0.
What i am doing wrong ?
NB- $a_1,a_2$ are components of $a$. O=(0,0)
What you are doing wrong depends upon the way how you deduced that that limit is equal to $0$. Anyway,\begin{align}\lim_{h\to0}\frac{f(ha_1,ha_2)}h&=\lim_{h\to0}\frac{h^3a_1{a_2}^3}{h^3{a_1}^3+h^6{a_2}^6}\\&=\lim_{h\to0}\frac{a_1{a_2}^3}{{a_1}^3+h^3{a_2}^6}\\&=\begin{cases}\frac{{a_2}^3}{{a_1}^2}&\text{ if }a_1,a_2\neq0\\0&\text{ otherwise.}\end{cases}\end{align}