Prove not differentiable at $(0,0)$ but directional derivatives every direction. Verify that the function $f: \mathbb R^2 \to \mathbb R$ given by $f(x,y) = \frac{x^2y}{x^4 + y^2}$ for $ (x,y) \neq 0$ and $f(0,0) = 0$ is not differentiable at $(0,0)$ and yet has directional derivatives in every direction.
My idea:
$\frac{\partial f(x,y)}{\partial x} = \frac{2xy(x^4+y^2)-4x^3(x^2y)}{(x^4+y^2)^2}$
$\frac{\partial f(x,y)}{\partial y} = \frac{x^2(x^4+y^2)-2y(x^2y)}{(x^4+y^2)^2}$
and neither of the partial derivatives $\frac{\partial f(x,y)}{\partial x}$ , $\frac{\partial f(x,y)}{\partial y}$ exist at $f(0,0)$, so by definition  they are not differentiable this point. 
Now let the unit vector $\hat{u}=(u_1,u_2)$ such that $\|\hat{u}\|=1$
Directional derivative at point $(0,0)$ is
$D_uf(0,0) = \lim_{t\to 0}\frac{f(0 +tu_1,0 +tu_2)-f(0,0)}{t} $ and must exist.
Is this right so far? Where do I go from here?

First of all:

$\frac{\partial f(x,y)}{\partial x} = \frac{2xy(x^4+y^2)-4x^3(x^2y)}{(x^4+y^2)^2}$
$\frac{\partial f(x,y)}{\partial y} = \frac{x^2(x^4+y^2)-2y(x^2y)}{(x^4+y^2)^2}$
and neither of the partial derivatives $\frac{\partial f(x,y)}{\partial x}$, $\frac{\partial f(x,y)}{\partial y}$ exist at $f(0,0)$

This is not right. The partial derivatives are particular directional derivatives (using the vectors $e_1=(1,0)$ and $e_2=(0,1)$), and you have to show that it exist (by definition below).
(As you wrote) Given a vector $u=(u_1,u_2)$ with $\|u\|=1$ we define the directional derivative of a function $f$ at a point $p=(p_1,p_2)$ by $$\dfrac{\partial f}{\partial u}(p) = \lim_{t\to 0} \dfrac{f(p+tu)-f(p)}{t} = \lim_{t \to 0} \dfrac{f(p_1+tu_1,p_2+tu_2) - f(p_1,p_2)}{t}$$ if the limit exist.
You have to show that this limit exist for every $u\in \Bbb{R}^2$ such that $\|u\|=1$.
Edit: To show that $f$ isn't differentiable at $(0,0)$, you can calculate the limit through different paths and conclude that the limits hasn't the same value, like Svetoslav said in comments.
