If all directional derivatives of $f$ in point $p$ exist, is $f$ differentiable?

Question

I am a little bit confused by the various theorems concerning the differentiability of a multivariable function.
Let $f : D \subseteq R^n \to R$ have all directional derivatives in point $p$. Does it directly imply that $f$ is differentiable in $p$? I know that the opposite is true: If $f$ were differentiable, it would imply that it has directional derivatives.

Answer 1

No, this is not true. Take, for instance$$\begin{array}{rccc}f\colon&\mathbb{R}^2&\longrightarrow&\mathbb{R}\\&(x,y)&\mapsto&\begin{cases}\frac{x^2y}{x^4+y^2}&\text{ if }(x,y)\neq(0,0)\\0&\text{ otherwise.}\end{cases}\end{array}$$You can check that, at $(0,0)$, every directional derivative exists. However, $f$ is not differentiable there.

Answer 2

Ler $$f(x,y)=\begin{cases} 1 & x^2< y < 2x^2\\ 0 & {\rm otherwise}\end{cases}$$ The directional derivatives are equal $0.$ The function is not continuous at $(0,0).$