You only need two things to prove this. First, the BAC-CAB rule:
$$A \times (B \times C) = B(A \cdot C) - C(A \cdot B)$$
And the product rule. Let $\dot \nabla \times (\dot F \times G)$ mean "differentiate $F$ only; pretend $G$ is constant here". So the product rule would read
$$\nabla \times (F \times G) = \dot \nabla \times (\dot F \times G) + \dot \nabla \times (F \times \dot G)$$
Now, apply the BAC-CAB rule. I'll do this for just one term for brevity:
$$\dot \nabla \times (\dot F \times G) = \dot F (\dot \nabla \cdot G) - G(\dot \nabla \cdot \dot F)$$
Now, here's where the dots become important: since $G$ is not differentiated in this whole equation, $\dot \nabla \cdot G$ is a directional derivative, conventionally written $G \cdot \nabla$. Indeed, we have
$$\dot F(\dot \nabla \cdot G) = (G \cdot \nabla) F$$
On the other hand, the $G(\dot \nabla \cdot \dot F)$ term can just drop the dots to get something that looks like a divergence:
$$G (\dot \nabla \cdot \dot F) = G(\nabla \cdot F)$$
Carry out the same expansion for the $\dot \nabla \times (F \cdot \dot G)$ term, and you're done.