Suppose $f, g \in L^{\infty}(\mathbb{R}^{2})$ and that $\partial_{x}f + \partial_{y}f=\partial_{x}g + \partial_{y}g=0$. I wish to evaluate the action
$$(\partial_{x}(fg) + \partial_{y}(fg), \phi) $$ where $\phi \in \mathcal{D}(\mathbb{R}^{2})$. In the smooth setting, of course we can say $\partial_{x}(fg) + \partial_{y}(fg) =0$ by the usual product rule. I am wondering if there is a way to justify this equality in the distributional sense?
Formally I would like to say $\partial_{x}(fg) + \partial_{y}(fg) = f(\partial_{x}g + \partial_{y}g) + g(\partial_{x}f + \partial_{y}f) = 0$ but not sure if we can justify the first equality. The second equality makes sense because $\partial_{x} g + \partial_{y}g = 0$ collapses into a smooth function (zero) so the product with $f$ makes sense (and similarly for $f$).
I suspect that it might not be possible since we cannot consistently define multiplication of continuous function with an arbitrary distribution.