If $\mathbf{F}$ and $\mathbf{G}$ are smooth and conservative. Find vector potential $\mathbf{H}$ for $\mathbf{F} \times \mathbf{G}$.
I tried to find it like this (kinda brute force-ishly)
$$\small \begin{align} \nabla \times \mathbf{H}&=\mathbf{F} \times \mathbf{G} \\ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ H_1 & H_2 & H_3 \end{vmatrix}&=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ F_1 & F_2 & F_3 \\ G_1 & G_2 & G_3 \end{vmatrix} \\ \mathbf{i}\left(\frac{\partial H_3}{\partial y}-\frac{\partial H_2}{\partial z}\right)-\mathbf{j}\left(\frac{\partial H_1}{\partial z}-\frac{\partial H_3}{\partial x}\right)+\mathbf{k}\left(\frac{\partial H_2}{\partial x}-\frac{\partial H_1}{\partial y}\right)&=\mathbf{i}\left(F_2 G_3 - F_3 G_2 \right)-\mathbf{j}\left(F_3 G_1 - F_1 G_3 \right)+\mathbf{k}\left(F_1 G_2 - F_2 G_1 \right)\end{align}$$ so $$\begin{cases} \frac{\partial H_3}{\partial y}-\frac{\partial H_2}{\partial z} = F_2 G_3 - F_3 G_2 \\ \frac{\partial H_1}{\partial z}-\frac{\partial H_3}{\partial x} = F_3 G_1 - F_1 G_3 \\ \frac{\partial H_2}{\partial x}-\frac{\partial H_1}{\partial y} =F_1 G_2 - F_2 G_1 \end{cases} $$ I don't see a simple way to solve $H_1$, $H_2$ and $H_3$.
Agh, yes, that's awfully painful. Here's something to try: Since $\mathbf F = \nabla f$ for some potential function $f$, what is $\nabla\times (f\mathbf G)$?
EDIT: So, the product rule tells us that $\nabla\times (f\mathbf G) = \nabla f \times \mathbf G + f\nabla\times\mathbf G$. Since $\mathbf G$ is conservative, $\nabla\times\mathbf G=\mathbf 0$, and so, setting $\mathbf H = f\mathbf G$, we have $\nabla\times\mathbf H = \nabla f\times\mathbf G = \mathbf F\times\mathbf G$, as desired.