Answer
$div (fF)=f[div F]+F \cdot \nabla f$
Work Step by Step
Show that $div (fF)=f[div F]+[F] \cdot [\nabla f]$
Consider $F=ai+b j+ck$
$div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$
Suppose, we have $F=F_1i+F_2j+F_3z$
Now, $div (fF)=\nabla \cdot (fF_1i+fF_2j+fF_3k)$
or, $=\dfrac{\partial [fF_1]}{\partial x}+\dfrac{\partial [fF_2]}{\partial y}+\dfrac{\partial [fF_3]}{\partial x}$
$div (fF)=f [\dfrac{\partial [F_1]}{\partial x}+\dfrac{\partial [F_2]}{\partial y}+\dfrac{\partial [F_3]}{\partial x}+[F_1i+F_2j+F_3z] \cdot [\dfrac{\partial f}{\partial x}i+\dfrac{\partial f}{\partial y}j+\dfrac{\partial f}{\partial z}k]$
This implies that
$div (fF)=f[div F]+[F] \cdot [\nabla f]$ (Proved)
