Answer
$div (fF)=f[div F]+F \cdot \nabla f$
Work Step by Step
When $F=A i+B j+C k$, then we have $div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$
Let us consider that $F=F_1i+F_2j+F_3z$
and $div (fF)=\nabla \cdot (fF_1i+fF_2j+fF_3k)=\dfrac{\partial [fF_1]}{\partial x}+\dfrac{\partial [fF_2]}{\partial y}+\dfrac{\partial [fF_3]}{\partial x}$
$=f [\dfrac{\partial [F_1]}{\partial x}+\dfrac{\partial [F_2]}{\partial y}+\dfrac{\partial [F_3]}{\partial x}+[F_1i+F_2j+F_3z] \cdot [\dfrac{\partial f}{\partial x}i+\dfrac{\partial f}{\partial y}j+\dfrac{\partial f}{\partial z}k]$
$=f[div F]+F \cdot \nabla f$
Hence, $div (fF)=f[div F]+F \cdot \nabla f$ (proved)
