If all directional derivatives exist at $0\in \mathbb R^2$, why the gradient doesn't exist? Let $f:\mathbb R^2\longrightarrow \mathbb R$ a function s.t. all directional derivative exist at $0$. Why the gradient doesn't exist ? 
Don't we have that $\nabla f(0)=(\frac{\partial f}{\partial x}(0),\frac{\partial f}{\partial y}(0))$ ? Indeed, $$\frac{\partial f}{\partial x}(0)=\nabla f(0)\cdot (1,0)\quad \text{and}\quad \frac{\partial f}{\partial y}(0)=\nabla f(0)\cdot (0,1)$$
so the gradient should exist, no ?
 A: As a counterexample, consider the function defined by
$$ f(r\cos\theta, r\sin\theta) = r\sin(3 \theta) $$
(in particular, $f(0,0)=0$ independently of which $\theta$ you choose).
This has all directional derivatives at $(0,0)$ -- actually $f$ restricted to any straight line through $0$ is linear -- but it cannot be approximated by a linear function near $0$. Therefore it doesn't have a gradient.

Your confusion may be that you think that the gradient is merely "whatever we get by making a vector out of the partial derivatives". However, a more abstract definition (which appears to be what your course is using) is

The gradient is the vector $w$ such that the directional derivative $D_v(f)$ equals $v\cdot w$ for all $v$.

We can easily see that if such a $w$ exists then its elements must be the partial derivatives -- but it is not guaranteed that $(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})$ has this property, and if it doesn't then it isn't a gradient.
In the example above, the partial derivatives are $(0,-1)$ -- but that doesn't work as $w$, because the directional derivative in the direction $(\cos \frac\pi6,\sin\frac\pi 6)$ is $1$, which does not equal $(0,-1)\cdot(\cos\frac\pi6,\sin\frac\pi6)=-\frac12$.
