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

Question

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...

2. 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.

Answer 1

1. The simplest answer is yes. Without getting too deep into the analysis, you can somewhat easily show that if it is not the case for some smooth path, then it is not the case for the derivative. That is, if the gradient is defined you get a derivative, and if not, not. Try the contrapositive for the limit of the derivative at the point if you are curious.

2. These are equivalent, specifically because you can have a derivative undefined along a certain direction UNLESS we define it there. If I define or find that the derivative along one of the axes exists, and the others are all continuous in the ball, they are continuous at the point and we have all the derivatives defined. This is neither weaker nor stronger, but directly equivalent. If the derivatives are discontinuous along other axes toward the point, we have an entire subspace approaching the point with possible discts or undefined derivative curves. (If this is incorrect, please someone else correct me!)