Don't linear combinations of partial derivatives entirely determine Jacobian? I'm learning multivariable calculus for the first time, having just taken a course on theoretical linear algebra and group theory, and the instructor motivated the Jacobian definition of a function derivative by giving an example of how, if we defined a multivariable derivative purely in terms of the existence of partial derivatives, we could get functions that are not continuous at a point, but for which the derivative still exists, which violates what we want from a definition of a derivative (as a stronger condition than continuity), and so we instead take the "existence of a linear map" definition instead.
But doesn't the existence of enough partial derivatives completely determine the Jacobian? For example, if you had $\frac{\partial f}{\partial x_i}$ for all components $0 \le i \le n$ for $f(x): \mathbb{R}^n \to \mathbb{R}$, don't you automatically guarantee the existence of the Jacobian
$$\begin{bmatrix} \frac{\partial f}{\partial x_1} \frac{\partial f}{\partial x_2} \cdots \frac{\partial f}{\partial x_n} \end{bmatrix}$$ by just using the partial derivatives you've shown exist. And does the existence of such a Jacobian satisfy the "existence of a linear map" condition to show that $f$ is differentiable, as a multivariable function?
My intuition is breaking down because I intuitively imagine a function as differentiable if the limits from all possible paths exist, and are the same. Aren't said limits entirely specified by these set of partial derivatives, from which you can get any direction derivative by a linear combination? And so in some sense, you've specified the datum for the "total derivative" (ie. $Df(x)$) when specifying all the partial derivatives?
 A: Existence of the Jacobian matrix at $c$ only needs, by definition, the partial derivatives at $c$. It is also true that a function is differentiable at $c$ if and only if the Jacobian exists at $c$ and $\lim_{\lvert h\rvert\to0}\frac{\left\lvert f(c+h)-f(c)-\nabla_c f\cdot h\right\rvert}{\lvert h\rvert}=0$.
However, it might still not be the case that, while the Jacobian at $c$ exists, $$\lim_{\lvert h\rvert\to0}\frac{\left\lvert f(c+h)-f(c)-\nabla_c f\cdot h\right\rvert}{\lvert h\rvert}=0$$
For instance, consider $$f(x,y)=\begin{cases}0&\text{if }x\le 0\lor y\le x^2\lor y\ge 4x^2\\ (x^2+y^2)^{1/4}\exp\left(\frac{1}{(y-x^2)(y-4x^2)}+\frac49x^{-4}\right)&\text{if }x>0\land x^2<y<4x^2\end{cases}$$
Then $f$ is $C^\infty$ on $\Bbb R^2\setminus\{0\}$, $\lim_{t\to 0} \frac{f(tv)-f(0)}{t}=0$ for all $v\in\Bbb R^2$, $f$ is continuous, but $f$ is not differentiable at $0$ because $$\limsup_{h\to 0}\frac{\lvert f(h)-f(0)-\nabla_0f\cdot h\rvert}{\lvert h\rvert}\ge\limsup_{t\to 0^+}\frac{\lvert f(t,5t^2/2)\rvert}{\lvert (t,5t^2/2)\rvert}=\limsup_{t\to 0^+}\left(t^2+\frac{25}{4}t^4\right)^{-1/4}=\infty$$
Also, $(x^2+y^2)^{-1/2}f(x,y)$ is again $C^\infty(\Bbb R^2\setminus\{0\})$, has all the directional derivatives equal to $0$ at $0$ and it isn't continuous at $0$.
Of course, by the total differential theorem, existence of the partial derivatives in a neighbourhood of $c$ and continuity of the partial derivatives at $c$ is sufficient to guarantee differentiability at $c$.
