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?

As you wrote:
$$D_uf(0,0) = \lim_{t\to 0}\frac{f(0 +tu_1,0 +tu_2)-f(0,0)}{t}=\lim_{t\to0}\frac{0-0}t=0$$
and thus the directional derivative at the origin exists at any direction (but the directon towards the origin itself, of course). 
