Consider a function $f \colon \mathbb{R}^2 \rightarrow \mathbb{R}$ that, as a distribution, has zero mixed partial derivative, that is, $ \langle \tfrac{\partial^2}{\partial x \partial y} f,\varphi \rangle = 0$ for any smooth test function $\varphi \colon \mathbb{R}^2 \rightarrow \mathbb{R}$ with compact support. Equivalently, $$ \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(x,y) \tfrac{\partial^2}{\partial x \partial y} \varphi(x,y) dx dy = 0.$$ If $f \in \mathcal{C}^2$ (twice continously differentiable), then $\tfrac{\partial^2}{\partial x \partial y} f(x,y) = 0$ everywhere. Integrating yields that, for all $x_1,x_2,y_1,y_2 \in \mathbb{R}$,
$$ f(x_2,y_2) - f(x_1,y_2) = f(x_2,y_1) - f(x_1,y_1). $$
Now, my question is whether the last equation remains to be true if $f \notin \mathcal{C}^2$.
Let $f^\epsilon = f*\eta^\epsilon$, where $\eta^\epsilon$ is the standard mollifier. Then $f^\epsilon$ is smooth and
$$f^\epsilon_{x_1x_2}(x) = \int_{\mathbb{R}^2} \eta^\epsilon_{x_1x_2}(x-y)f(y)\, dy = 0,$$
since you can treat $\eta^\epsilon_{x_1x_2}(x-\cdot)$ as a test function. So your identity holds for $f^\epsilon$, and if $f$ is continuous then $f^\epsilon \to f$ uniformly as $\epsilon \to 0$ and so the identity holds for $f$.