Let $v \in H^1(\mathbb{R}^2)^2$ such that $\operatorname{div} \ v = 0$ and $\omega = \operatorname{curl} \ v$, where $\operatorname{curl}$ is defined as $\operatorname{curl} \ v = \partial_1 v_2 - \partial_2 v_1$. Show that $\operatorname{curl}\ ((v \cdot \nabla)v) = v \cdot \nabla \omega$ in the sense of distributions. (The operator $(v \cdot \nabla)$ is defined as $\sum_i v_ i \partial_i$).
I really don't know where to start. The only hint I have is to start with functions $v \in \mathcal{C}^2(\mathbb{R^2})$.
I think the goal is to show that for every test function $\varphi \in \mathcal{D}(\mathbb{R}^2)$ (smooth compactly supported functions) we have that $$ \int \operatorname{curl} \ ((v \cdot \nabla)v) \varphi = \int v \cdot \nabla \omega \varphi. $$
Usually in questions like this my first idea is to integrate by parts, but I don't know how to apply that here.
Edit: The part for $\mathcal{C}^2(\mathbb{R}^2)$ was actually pretty easy, I just needed to write everything in terms of the coordinates and identify the terms by swapping $\partial_1$ and $\partial_2$ which is allowed by Schwarz's Theorem. However I still don't know how to transfer this to the Sobolev space.