Answer
$\nabla^2(fg)=f \nabla^2 g+g \nabla^2 f+2 \nabla f\cdot \nabla g$
Work Step by Step
We need to show that $\nabla^2(fg)=f \nabla^2 g+g \nabla^2 f+2 \nabla f\cdot \nabla g$
Here, $f$ and $g$ are twice differentiable functions.
We need to use $\nabla(fg)=f \nabla g+g \nabla f $ and $\nabla(f+g)=\nabla f+\nabla g $
Thus, we have$\nabla^2(fg)=\nabla (f \nabla g+g \nabla f )=\nabla (f \nabla g)+\nabla (g \nabla f)$
This implies that
$\nabla (f \nabla g)+\nabla (g \nabla f)=[f \nabla^2 g+(\nabla f)(\nabla g)]+[g \nabla^2 f+(\nabla g)(\nabla f)]$
or, $\nabla^2(fg)=f \nabla^2 g+g \nabla^2 f+2 \nabla f\cdot \nabla g$
Hence, the result has been proved.