Q:

Equal partial derivatives implies symmetry

The statement reads as follows:
Let $f$ be continuously differentiable if $\frac{\partial f}{\partial x}(x,y)=\frac{\partial f}{\partial y}(x,y)$ then $f(a,-a)=f(-a,a)$
I would like to know if this statement is true or false. Intuitively it seems possible since having the same slope from either direction implies a sort of symmetry. Thus I think this statement is true. Is there a way to prove/falsify it?

A:

Yes, this is true.
Let $g(t) = f(t,-t)$. Then $g\in C^1(\mathbb{R})$, and
$$ g'(t) = \frac{\partial f}{\partial x}(t,-t)\frac{dt}{dt} + \frac{\partial f}{\partial y}(t,-t)\frac{d(-t)}{dt} = \left(\frac{\partial f}{\partial x}-\frac{\partial f}{\partial y}\right)(t,-t) = 0.
$$
Hence, $g$ is constant. In particular, $f(a,-a) = g(a) = g(-a) = f(-a,a)$.

