Find the directional derivatives of a function 
Consider the function $f : R^2 → R$ given by 
$$
f(x, y) =
\begin{cases}
\begin{matrix}
\frac{x^2y}{x^4+y^2} & \mathrm{if}\ (x, y) \ne(0, 0)\\
0 & \mathrm{if}\ (x, y) = (0, 0) \\
\end{matrix}
\end{cases}
$$
Using the definition, compute the directional derivative $\partial_uf(0,0) $ for
  all directions $u=(u_1, u_2)\ne (0, 0)$. [Hint: Consider the cases $u_2 \ne 0$
  and $u_2 = 0$ separately, and use that $\partial _uf(0, 0) = \lim_{h\to 0} \frac{f(hu_1,hu_2)−f(0,0)}{h}$.]

Here is my solution:
We know that $f(0,0)=0$, we can rewrite the formula for the directional derivative as $$\partial _uf(0, 0) = \lim_{h\to 0} \frac{f(hu_1,hu_2)}{h}$$
We have that in the function, $x$ is denoted by $hu_1$ and $y$ is denoted by $hu_2$.
We now look at each case, in terms of the original function.
Case 1: We have that $u_2=0$,
$$\lim_{h\to0}\left(\frac{((hu_1)^2(h\cdot 0))}{h ((hu_1)^4+(hu_2)^2)}\right) = 0$$
Case 2: We have that $u_2\ne0$,
$$\lim_{h\to0}\left(\frac{((hu_1)^2(hu_2))}{h ((hu_1)^4+(hu_2)^2)}\right) = \lim_{h\to0}\left(\frac{h^3(u_1^2u_2)}{h^3(h^2 u_1^4+u_2^2)}\right) = \lim_{h\to0}\left(\frac{u_1^2u_2}{h^2 u_1^4+u_2^2}\right)$$
Solving as $h\to0$ we see that the $\lim\to \frac{u_1^2}{u_2}$.
Therefore the directional derivatives for $\partial_uf(0,0)$ in the direction $u=(u_1,0)$ is $0$. The directional derivative in the direction $u=(u_1,u_2)$ is $\frac{u_1^2}{u_2}$.
Is this solution correct? How can I improve this answer in general?

There are some typos ! The quotient $\frac{f(hu_1,hu_2)}{h}$ is given by
$$\frac{f(hu_1,hu_2)}{h}= \frac{u_1^2u_2}{h^2u_1^4+u_2^2}.$$
If $u_2=0$, then $\frac{f(hu_1,hu_2)}{h}=0 \to 0$ as $h \to 0.$
If $u_2 \ne  0$, then $\frac{f(hu_1,hu_2)}{h} \to \frac{u_1^2}{u_2}$ as $h \to 0.$
