Curl of Cross Product of Two Vectors I want to prove the following identity

$$\text{curl } \left(\textbf{F}\times \textbf{G}\right) = \textbf{F}\text{ div}\textbf{ G}- \textbf{G}\text{ div}\textbf{ F}+ \left(\textbf{G}\cdot \nabla \right)\textbf{F}- \left(\textbf{F}\cdot \nabla \right)\textbf{G}$$

But I do not know how! Also, what does $\textbf{F}\cdot \nabla $ mean, isn't it the divergence of $\textbf{F}$!
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.
It's easiest to see by writing it out in components:

[ ( ∇ ⋅ a ) b ]i = ( ∂x ax + ∂y ay + ∂z az )bi = ( ∂x ax )bi + ( ∂y ay )bi + ( ∂z az )bi

Whereas 

[(∇⋅a)b]i=(∂x ax +∂y ay +∂z az)bi =(∂x ax)bi +(∂y ay)bi +(∂z az)bi 

and clearly these are not the same. So while a ⋅ b = b ⋅ a a⋅b=b⋅a holds when a and b are really vectors, it is not necessarily true when one of them is a vector operator. This is one of the cases where the convenience of considering ∇ ∇ as a vector satisfying all the rules for vectors does not apply.