Let $D \subset \mathbb{R}^d$ be a smooth bounded domain. Let $C_c^\infty(D)$ denote smooth and compactly supported functions on $D$.
Let $f \in [C_c^\infty(D)]^d$ be a smooth, compactly supported vector field on $D$, such that $$\int_D f \cdot \nabla \phi = 0 \quad \forall \phi \in C_c^\infty(D).$$
Question: What can be said about $f$, besides that $\mathrm{div} \, f = 0$?
Edit: If $d = 3$ then $f$ could be $f = \mathrm{curl} \, w$ for any $w \in [C_c^\infty(D)]^3$, so not much.
The only answer is: absolutely nothing.
Take the reverse, i.e. let a locally integrable vector field $g$ satisfy the indentity $$ \int\limits_Dg\cdot f\,dx=0\quad\forall\, f\in \bigl[C_c^{\infty}(D)\bigr]^d\colon\,{\rm div}\,f=0.$$ Such vector field $g$ is known to be potential. More precisely, there is a locally weakly differentiable function $\phi$ such that $g=\nabla\phi\,$ a.e. in $D$. The reverse side of your question is what else can be said about $g$, besides that $g=\nabla\phi\,$? While the answer stays the same, absolutely nothing.