Let $F$ be a conservative vector field (smooth and defined on all $\mathbb R^2$), that is, $\int_P F \cdot dr$ depends only on the start and end of path $P$. Then at every point $\frac {\partial F_y}{\partial x} = \frac {\partial F_x}{\partial y}$.
To better understand this, I developed the proof below, to which I request verification, feedback, and comments. Note that there may be faster ways to prove this, drawing on preexisting results, but my goal is to build intuition, not find the fastest proof.
Consider a rectangular region from $(0,0)$ to $(u,v)$, and the paths along its edges: $$\begin{align} B &= (0,0)...(u,0) \\ T &= (0,v)...(u,v) \\ L &= (0,0)...(0,v) \\ R &= (u,0)...(u,v). \\ \end{align}$$
Then, by the MVT, the work along each path is: $$\begin{align} W_B &= u F_{B_x} \\ W_T &= u F_{T_x} \\ W_L &= v F_{L_y} \\ W_R &= v F_{R_y} \end{align}$$ where $F_B, F_T, F_L, F_R$ are the value of $F$ at some particular point along each respective path (with $F_{B_x}$ meaning the $x$ component of $F_B$).
Then $$W_L + W_T - W_R = W_B \\ % W_L - W_R = W_B - W_T \\ v (F_{L_y} - F_{R_y}) = u (F_{B_x} - F_{T_x}).$$
Since $F$ is smooth, as the region is made arbitrarily small (i.e. $u \to 0, v \to 0$) [*], there exists in the region a point $p$ where $$\begin{align*} \frac {\partial F_y}{\partial x} &\to \frac {F_{R_y} - F_{L_y}} u \\ \frac {\partial F_x}{\partial y} &\to \frac {F_{T_x} - F_{B_x}} v \\ \end{align*}$$ giving $$\frac {\partial F_y}{\partial x} \to \frac {\partial F_x}{\partial y}.$$ Since we can construct and shrink such a region around any point, we have, for all points $$\frac {\partial F_y}{\partial x} = \frac {\partial F_x}{\partial y}.$$
[*] Question: Is the statement marked [*] justified? I'm concerned that in general the $x$ component of the point along $F_L$ does not necessarily equal the $x$ component of the corresponding point along $F_R$. So I can't simply apply the one dimensional MVT. However, I believe the argument is still correct, and that the MVT can be used this way as well. Am I correct? If so: How do you support that? If not: Is there an alternate argument, or "fix" of the proof?
Even better: How can I revise this proof so there's no ambiguity about its justification?