Does the Existence of Derivatives along All Smooth Paths Guarantee Multivariable Differentiability?

This question is inspired by other relevant questions on MSE regarding continuity of partial derivatives and differentiability, existence of partial derivatives and differentiability, and limits along arbitrary smooth curves.

I'm trying to get a handle on the extra bit of weirdness introduced by multivariable functions when it comes to differentiability and continuity. The applicable definition here is:

Definition (Cain & Herod, 1997): Let $$f: D \rightarrow \mathbb{R}^p$$, where $$D \subset \mathbb{R}^n$$ and let $$\mathbf{x}$$ be an interior point of $$D$$. Then $$f$$ is differentiable if there exists a linear function $$L$$ such that $$\lim\limits_{\mathbf{h} \rightarrow \mathbf{0}} \dfrac{\|f(\mathbf{x} + \mathbf{h}) - f(\mathbf{x}) - L(\mathbf{h})\|}{\|\mathbf{h}\|} = \mathbf{0}$$

So, given that the multivariable limit in the definition above requires that the limit exists (and is zero) independent of the path, can I interpret this as translating to the requirement that the derivative along any smooth path through $$\mathbf{x}$$ must exist?

That is, if we define a so-called path derivative $$D_\mathbf{p}f$$:

$$D_{\mathbf{p}}f : \mathbf{x} \mapsto \lim\limits_{t \rightarrow t_0} \dfrac{f(\mathbf{x} + \mathbf{p}(t)) - f(\mathbf{x})}{\|\mathbf{p}(t)\|}$$

where $$\mathbf{p}$$ defines any path with $$t_0$$ such that $$\lim\limits_{t \rightarrow t_0} \mathbf{p}(t) = \mathbf{0}$$,

1. If a path is defined as smooth if it is representable by an infinitely differentiable vector function, is differentiability equivalent to the existence of all path derivatives where $$\mathbf{p}$$ defines a smooth path?

Reading this answer, I initially suspected that this path derivative condition was identical to the total derivative which defines differentiability, but I wonder what kind of wrinkle the requirement that $$\mathbf{p}$$ defines a smooth path introduces. Proving that differentiability implies the existence of all such path derivatives is straightforward enough. I can't see if there's a way to prove differentiability assuming the existence of all path derivatives, however...

1. How does this condition compare to the condition of the existence of all partial derivatives and $$n-1$$ continuous partial derivatives in some $$n$$-ball containing $$\mathbf{x}$$ (quoted here)? Is it stronger/weaker?

At the very least, I think that this condition is not useless. For example, if you can find a smooth path for which the path derivative does not exist, then $$f$$ is not differentiable. In reference to this question–where one partial derivative is not defined along some line with $$\mathbf{x}$$ deleted–the notion of path derivatives implies that such a function is automatically not differentiable.

• The existence of the path derivatives does not seem to imply that there is a linear functional ... take for example $|x|$ in dimension $1$. Aug 22, 2021 at 2:08
• What is your definition of "smooth path"? Aug 27, 2021 at 23:20
• @PaulFrost A path which can be represented by an infinitely differentiable vector function.
– Ugo
Aug 27, 2021 at 23:41
• So the path derivative (if it exists) is always a real number. I doubt that this is a reasonable concept. Sep 1, 2021 at 17:27
• @Ugo Yes, in order to give you something reasonable it should contain the partial derivatives as special cases. Sep 6, 2021 at 16:05