Answer
curl $(F\times G)=F (div G)-G (div F)+(G \cdot \nabla)F-(F \cdot \nabla)G$
Work Step by Step
Use definition of curl $(F\times G)=\nabla \times (F\times G)$
We need to use the product rule:
curl $(F\times G)=\dot{\nabla}\times (\dot{F}\times G)+\dot{\nabla}\times (F\times \dot{G})$
Also, we have $u \times (v \times w)=v(u \cdot w)-w(u \cdot v)$
Now, curl gives: $(F\times G)=[F(\dot{\nabla} \cdot G)-G(\dot{\nabla} \cdot \dot{F})]+[F(\dot{\nabla} \cdot \dot{G})-\dot {G}(\dot{\nabla} \cdot F)]=F (div G)-G (div F)+\dot {F}(\dot{\nabla} \cdot G)-\dot {G}(\dot{\nabla} \cdot F)$
This implies that
curl $(F\times G)=F (div G)-G (div F)+(G \cdot \nabla)F-(F \cdot \nabla)G$
Hence, it has been proved.