Curl Proof Question Prove the given formula. 

So far I have $f\textbf{F}=(f\textbf{F}_1, f\textbf{F}_2, f\textbf{F}_3)$, but I'm not sure where to go from there. Could anyone give me some pointers? 
Thank you. 
 A: Now compute the curl of that, being sure to apply the product rule properly.
It may also help if you compute the other side of the identity also, and then reorganize stuff to make things match.

Here is a foothold for you to start with. By definition, $\mathrm{curl}(fF_1,fF_2,fF_3):=\left(\left(\frac{\partial fF_3}{\partial y}  - \frac{\partial fF_2}{\partial z}\right),\left(\frac{\partial fF_1}{\partial z} - \frac{\partial fF_3}{\partial x}\right) , \left(\frac{\partial fF_2}{\partial x} - \frac{\partial fF_1}{\partial y}\right) \right)$.
Each of the partial derivatives here requires you to take the derivative of a product of functions.

To add a little more detail, let's do the first term. The second line is obtained by the first by simply applying the product rule.
$$\frac{\partial fF_3}{\partial y}  - \frac{\partial fF_2}{\partial z}=\\
\frac{\partial f}{\partial y}F_3+\frac{\partial F_3}{\partial y}f  - \left(\frac{\partial f}{\partial z}F_2+\frac{\partial F_2}{\partial z}f\right)=\\ \frac{\partial F_3}{\partial y}f-\frac{\partial F_2}{\partial z}f+ \frac{\partial f}{\partial y}F_3-\frac{\partial f}{\partial z}F_2=\\ f\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right) + \left(\frac{\partial f}{\partial y}F_3-\frac{\partial f}{\partial z}F_2\right)
$$
Notice that the left hand term is the $x$ component of $f\cdot\mathrm{curl}(F)$ and the right half is the $x$ component of $(\nabla f)\times F$.
The other two pieces proceed exactly the same way.
