If a scalar field $f$ is differentiable at a point $\mathbf{a}$ in $\mathbb{R}^n$, then the directional derivative $f^\prime( \mathbf{a}; \mathbf{y})$ exists for every direction $\mathbf{y}$, and we have $$ f^\prime( \mathbf{a}; \mathbf{y} ) = \nabla f (\mathbf{a}) \cdot \mathbf{y}. $$
Now my question is, does this formula always hold as long as all the partial derivatives of $f$ exist at the point $\mathbf{a}$? If so, then how to prove this? If not, then what counter-example(s) can we give?
P.S.:
Suppose that $f$ is a scalar field defined on a set $S$ in $\mathbb{R}^n$, let $\mathbf{a}$ be an interior point of $S$, and let $\mathbf{y}$ be any point in $\mathbb{R}^n$ such that all the partial derivatives of $f$ exist at $\mathbf{a}$ and the directional derivative $f^\prime ( \mathbf{a}; \mathbf{y} )$ exist. Now my question is, does the formula $$ f^\prime ( \mathbf{a}; \mathbf{y} ) = \del f( \mathbf{a}) \cdot \mathbf{a} $$ always hold?
Of course, this formula does hold when $f$ is differentiable at $\mathbf{a}$. This much I know.
Define $f(x,y)=0$ on the $x,y$ axes, and $f(x,y)=x$ everywhere else. Then both partial derivatives of $f$ exist and equal $0$ at $(0,0).$ Thus $\nabla f (0,0) = (0,0).$ But the directional derivatives of $f$ exist and are nonzero in any other direction. Thus the formula fails in this situation.