Example for a continuous function that has directional derivative at every point but not differentiable at the origin Can we find a function $f:\mathbb R^n\to\mathbb R$ that such that $f$ is continuous and $\partial_v f(p)$ exists for all $p\in\mathbb R^n$ and $v\in\mathbb R^n$. But $f$ is not differentiable at $0$?
Is such function $f$ exists?
Here give a example that has directional derivative everywhere, but it's not continuous at the origin.
 A: Consider the polar coordinates $\ (r\,\ \Phi)\ $ in $\ \mathbb R^2.\ $ Function $ f:\mathbb R^2\rightarrow\mathbb R,\ $ which in polar coordinates is given by:
$$ f(r\,\ \Phi)\,\ :=\,\ r\cdot\sin(3\cdot\Phi) $$
is continuous everywhere, is infinitely differentiable  outside the origin $\ O,\ $ and $\ f\ $ has all directional derivative at $\ O,\ $ but $\ f\ $ is not differentiable at $\ O.$
A: On $\mathbb R^2$ define
$$ f(x,y) = \begin{cases} \dfrac{xy^2}{x^2+y^2}& (x,y)\ne (0,0) \\ 0 & (x,y)= (0,0)\end{cases}$$
Then $f\in C^\infty(\mathbb R^2 \setminus \{(0,0)\}),$ and $f$ is continuous at $(0,0).$ For any unit vector $u= (a,b),$ we have
$$\tag 1 \frac{f(tu)}{t} = ab^2.$$
Thus $f$ has directional derivatives at $(0,0)$ in all directions.
Suppose $Df(0,0)$ exists. Because $f$ is $0$ on the axes, all partial derivatives of $f$ at $(0,0)$ vanish, so $Df(0,0)$ must be the $0$ linear transformation. Thus $f(x,y) = o((x^2+y^2)^{1/2}).$  But that violates $(1),$ so we have a contradiction. This proves $f$ is not differentiable at $(0,0).$
