Is the converse of the statement "If partial derivatives exist and are continuous, then the function is differentiable at that point" true?

Question

I know the proof of the statement "If partial derivatives exist in a n-ball $B(\vec{a})$ and are continuous at $\vec{a}$, then the function is differentiable at $\vec{a}$ " . But I was wondering whether the converse holds.

I was thinking along these lines that, differentiability at a point implies the existence of tangent plane, and that $\vec{a}$ lies on that plane. This would mean that the partial derivatives do exist, but I'm not able to visualize or proceed any further.

Answer 1

I don't have repututation for comment so I will post as answer. The converse of your claim is not true even in dimension 1: differentiability in one point does not imply diffenentiability in any neighborhood of that point, see for example:

\begin{cases}x^2 \, \,\,\text{ if} &x \in \mathbb{Q} \\ 0 &\text{otherwise} \end{cases} It is differentiable at the origin and not continuous in any other point. The same example works also in higher dimension.