The partial derivatives at $(0,0)$ are
$$\partial_x f(0,0) = \lim_{h\to 0} \frac{f(h,0) - f(0,0)}h = 0$$
$$\partial_y f(0,0) = \lim_{h\to 0} \frac{f(0,h) - f(0,0)}h = 0$$
so the only candidate for the differential $Df(0,0)$ is the zero operator.
However, the limit
$$\lim_{(h_1, h_2) \to (0,0)} \frac{\|f(h_1, h_2) - f(0,0) - Df(0,0)(h_1, h_2)\|}{\|(h_1, h_2)\|} = \lim_{(h_1, h_2) \to (0,0)} \frac{\|f(h_1, h_2)\|}{\sqrt{h_1^2 + h_2^2}} = \lim_{(h_1, h_2) \to (0,0)} \frac{h_1h_2^2}{(h_1^2 + h_2^2)^{3/2}}$$
does not exist because e.g. for $h_1 = h_2$ we get $$\lim_{h_1 \to 0} \frac{h_1^3}{|h_1|^3}$$
which is $\pm 1$, depending on whether $h_1$ approaches $0$ from the left or from the right.
Therefore, $f$ is not differentiable at $(0,0)$.
Take the directional derivative of $f$ along the direction $(1,1)$. For all points of the form $(t,t)$, we have
$$\frac{f(t,t) - f(0,0)}{t} = \frac{1 - \sqrt{\vert t^2 \vert} - 1}{t} = \frac{- \vert t \vert}{t}$$
The limit as $t \rightarrow 0$ does not exist because:
Take $t = + \epsilon$ where $\epsilon >0$, the limit is $-1$ as $\epsilon \rightarrow 0$. Now, take $t = -\epsilon$ where $\epsilon >0$ and let $\epsilon \rightarrow 0$, the limit will be $1$. So two different limits while you're going to zero. Hence, limit doesn't exist.
Best Answer
It's easy. You only have to see that if $f$ is differentiable in $(0,0)$, then $f'(0,0)$ is a linear transformation. So:
$$f'(0,0)\cdot (1,1)=f'(0,0)\cdot((1,0)+(0,1))=f'(0,0)\cdot(1,0)+f'(0,0)\cdot(0,1).$$
And that is a contradiction, because $f'(0,0)\cdot (1,1)\not = f'(0,0)\cdot(1,0)+f'(0,0)\cdot(0,1).$