I have attached the Question.
Im not sure how to prove this identity, I was thinking maybe giving arbitary vectors?
If $\displaystyle \phi$ is a function of x, y and z, and I denote $\displaystyle \phi_x$ for $\displaystyle \tfrac{\partial \phi}{\partial x}$ (similar for y and z), then
$\displaystyle \phi \nabla \phi = \phi(\phi_x,\phi_y,\phi_z) = (\phi\phi_x,\phi\phi_y,\phi\phi_z)$
Now all you have to do is calculate the curl of this vector (i.e. nabla 'cross product' this vector); but remember that both $\displaystyle \phi$ and the three partial derivatives are functions of x, y and z.