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