How do I solve $F = \nabla\times G$ for $G$?

The problem amounts to finding a vector potential $\mathbf G$ such that $\mathbf F = \nabla \times \mathbf G$. The existence of a vector potential for $\mathbf F$ is equivalent to saying that $\mathbf F$ is divergence free: $$\nabla \cdot \mathbf F = \nabla \cdot (\nabla \times \mathbf G) = 0.$$

In our case we have $\nabla \cdot \mathbf F = 0$, so there indeed exists a vector potential $\mathbf G$.

Now note that the choice for $\mathbf G$ is far from unique. Let $f$ be any differentiable scalar function on $\mathbb R^3$. We then have $$\nabla \times(\mathbf G + \nabla f) = \nabla \times \mathbf G .$$ Thus $\mathbf G' = \mathbf G + \nabla f$ is also a vector potential for $\mathbf F$. We can exploit this non-uniqueness to simplify the set of equations $$ \partial_y G_z - \partial_z G_y = F_1, \quad \partial_z G_x - \partial_x G_z = F_2, \quad \partial_x G_y - \partial_y G_x = F_3 \; ,$$ by choosing our $f$ such that $\partial_z f = -G_z$. This simplifies the set of equations to $$-\partial_z G'_y = F_1, \quad \partial_z G'_x = F_2, \quad \partial_x G'_y - \partial_y G'_x = F_3 \;.$$ This choice of $f$ is purely to start off with a simple set of equations, and does not need calculation.

The problem now amounts to finding $\mathbf G'$. Integrating the first two equations yields $G'_y$ and $G'_x$ with integration constants $C_1(x,y)$ and $C_2(x,y)$ respectively. In the last equation we have to make a choice for these constants.