There are no directional derivatives in nearly all directions. Consider, in particular, along the line $y=x$. $f(x,y)$ is a constant times the absolute value function.
When a function of two variables is differentiable, then there is a tangent plane to the surface $z=f(x,y)$, and there are directional derivatives in all directions. This one doesn't have directional derivatives except in two directions, and there's no tangent plane to the surface $z=f(x,y)$.
It's easy to prove that $\lim_{(x,y)\to (0,0)}f(x,y)=0=f(0,0)$ (see this) hence $f(x,y)$ is continous at $(0,0)$. To determine if $f(x,y)$ is differentiable, we need the values of the derivatives $f_{x}(0,0)$ and $f_{y}(0,0)$.
By definition of partial derivatives:
$\\ f_{x}(0,0)=\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to 0}0=0 \\ \\ f_{y}(0,0)=\lim_{k\to 0}\frac{f(0,k)-f(0,0)}{k}=\lim_{k\to 0}0=0$
Now $f(x,y)$ is differentiable in $(0,0)$ if and only if $$\lim_{(h, k)\to (0,0)}\frac{f(h,k)-f(0,0)-f_{x}(0,0)k-f_{y}(0,0)k}{\sqrt{h^2+k^2}}=0$$
which becomes
$$\lim_{(h, k)\to (0,0)}\frac{h k^3}{(h^2+k^4)\sqrt{h^2+k^2}}=0$$
but if you take $h=k^2$, the restriction is $$\lim_{k\to 0}\frac{k}{2\sqrt{k^2+k^4}}=\lim_{k\to 0}\frac{k}{2|k|\sqrt{1+k^2}}=\begin{cases}\frac{1}{2}&\mbox{if} \ k\to 0^{+}\\ -\frac{1}{2}&\mbox{if}\ k\to 0^{-}\end{cases}$$
On the path $(k^2,k)$ the limit
$$\lim_{(h, k)\to (0,0)}\frac{h k^3}{(h^2+k^4)\sqrt{h^2+k^2}}\ne 0$$ hence $f(x,y)$ is not differentiable in $(0,0)$.
Best Answer
To show that $f$ is differentiable at $(0,0)$ you have to show that $$f(h)=f(0,0)+\nabla f (0,0) \cdot h + o(|h|)$$ for $h \in \Bbb{R}^2$ in a neighbourhood of $(0,0)$ (here $\cdot$ denotes the scalar product). It is natural to put $\nabla f (0,0) = (0,0)$, so that indeed you need to prove $$\lim_{h \to (0,0)} \frac{f(h)-f(0,0)}{|h|} =0$$ Using polar coordinates $h=(R \cos \theta , R \sin \theta)$ you have $$0 \le \frac{f(h)-f(0,0)}{|h|} = \frac{R^4 \cos^2 \theta \sin^2 \theta}{R^2} \le R^2 \to 0$$ and you are done.