$\def\RR{\mathbb{R}}$ $\def\bv{\textbf{v}}$ $\def\bw{\textbf{w}}$ $\def\bu{\textbf{u}}$ $\def\bx{\textbf{x}}$ $\def\a{\alpha}$ $\def\D{\Delta}$
Below are two proofs of the symmetry of second partial derivatives. I merely wish verification that the first proof is valid, whilst the second is not.
Theorem: Let $f$ be a function $\RR^n\to \RR^m$. If $\partial_{\bv}\partial_{\bw} f$ and $\partial_\bw\partial_\bv f$ exist and are continuous on some open set $V'$ (for some $\bv, \bw\in \RR^n$), then $$\partial_\bv\partial_\bw f = \partial_\bw\partial_\bv f \ \text{(on $V'$)}.$$
It is sufficient to prove the result for a function $\RR^n\to\RR$, so both arguments below restrict themselves to that case. For a function $g:\RR^n\to \RR$, a vector $\bu\in\RR^n$ and a nonzero scalar $h\in\RR$, we let $$\D_{\bu,h}g(\bx) := \frac{g(\bx+h\bu)-g(\bx)}{h}.$$
Proof #1: The sketch of this proof is as follows: $$\partial_\bv\partial_\bw f(\bx) = \lim_{t\to 0}\D_{\bw,t}\D_{\bv,t}f(\bx) = \lim_{t\to 0}\D_{\bv,t}\D_{\bw,t}f(\bx) = \partial_\bw\partial_\bv f(\bx).$$ The second equality is trivial. We prove the first equality (the third is shown similarly). We wish to move freely in the $\bv$ and $\bw$ directions, so choose $t$ such that $\bx + t_1\bv + t_2\bw \in V'$ for all $t_1,t_2\in[0,t]$. Now observe that \begin{equation} \begin{split} \D_{\bv,t}\D_{\bw,t}f(\bx) = \partial_\bv \D_{\bw,t}f(\bx+\a_\bv t\bv) = \D_{\bw,t}\partial_\bv f(\bx+\a_\bv t\bv) = \partial_\bw\partial_\bv f(\bx+\a_\bv t\bv + \a_\bw t\bw) \end{split} \end{equation} for $\a_\bv,\a_\bw\in (0,1)$, where both equalities use the Mean Value Theorem. Taking the limit as $t\to 0$ gives the desired result. $$\blacksquare$$
The switch from $\partial_\bv \D_{\bw,t}f$ to $\D_{\bw,t}\partial_\bv f$ may seem unnecessary, so I was tempted to simplify the proof as follows:
Proof #2: The sketch now is as follows: $$\partial_\bv\partial_\bw f(\bx) = \lim_{t\to 0}\D_{\bv,t}\D_{\bw,t}f(\bx) = \lim_{t\to 0}\D_{\bw,t}\D_{\bv,t}f(\bx) = \partial_\bw\partial_\bv f(\bx).$$ As before, the second equality is trivial. We prove the first equality (the third is shown similarly). We wish to move freely in the $\bv$ and $\bw$ directions, so choose $t$ such that $\bx + t_1\bv + t_2\bw \in V'$ for all $t_1,t_2\in[0,t]$. Now observe that \begin{equation} \begin{split} \D_{\bv,t}\D_{\bw,t}f(\bx) = \D_{\bv,t} \partial_{\bw,t}f(\bx+\a_\bw t\bw) = \partial_\bv\partial_\bw f(\bx+\a_\bv t\bv+\a_\bw t\bw) \end{split} \end{equation} for $\a_\bv,\a_\bw\in (0,1)$, where again both equalities use the Mean Value Theorem. Taking the limit as $t\to 0$ gives the desired result. $$\blacksquare$$
Question: proof #1 is valid, while proof #2 is not, right?
The reason as to why I believe the second proof is not valid is that when writing $$\D_{\bv,t}\D_{\bw,t}f(\bx) = \D_{\bv,t} \partial_{\bw,t}f(\bx+\a_\bw t\bw)$$ the $\a_\bw$ cannot be taken to be a constant, but rather it must depend on $\bx$. In particular
$$\D_{\bv,t}\D_{\bw,t}f(\bx) = \frac{1}{t}\left[\frac{1}{t}\Big(f(\bx+t\bv+t\bw)-f(\bx+t\bv)\Big) - \frac{1}{t}\Big(f(\bx+t\bw)-f(\bx)\Big)\right]$$ $$ \ \ = \frac{1}{t}\Big[\partial_\bw f(\bx+\a_\bw(\bx+t\bv)t\bw) - \partial_\bw f(\bx+\a_\bw(\bx)t\bw)\Big]$$ where $\a_\bw(\bx+t\bv)$ need not equal $\a_\bw(\bx)$. From here on the proof falls apart.
The first proof does seem to avoid such mistake: letting $g:\bx\mapsto \D_{\bw,t}f(\bx)$ we get that $g$ is differentiable with respect to $\bv$, with $$(\partial_\bv g)(\bx) = \frac{\partial_\bv f(\bx + t\bw) - \partial_\bv f(\bx)}{t} = (\D_{\bw,t}(\partial_\bv f))(\bx)$$ justifying the switch from $\partial_\bv \D_{\bw,t}f$ to $\D_{\bw,t}\partial_\bv f$. Furthermore, the $\a_\bv, \a_\bw$ are indeed constants since $$(\D_{\bv,t}g)(\bx) = \frac{g(\bx+t\bv)-g(\bx)}{t} = (\partial_\bv g)(\bx+\a_\bv t\bv)$$ for some fixed $\a_\bv\in (0,1)$ and so on.