Answer
$$\operatorname{div}(f \mathbf{F})=f \operatorname{div}(\mathbf{F})+\mathbf{F} \cdot \nabla f$$
Work Step by Step
We verify as follows:
\begin{align*}
\operatorname{div}(f \mathbf{F})&=\nabla \cdot(f \mathbf{F})\\
&=\frac{\partial}{\partial x}\left(f F_{1}\right)+\frac{\partial}{\partial y}\left(f F_{2}\right)+\frac{\partial}{\partial z}\left(f F_{3}\right)\\
&=\frac{\partial f}{\partial x} F_{1}+f \frac{\partial F_{1}}{\partial x}+\frac{\partial f}{\partial y} F_{2}+f \frac{\partial F_{2}}{\partial y}+\frac{\partial f}{\partial z} F_{3}+f \frac{\partial F_{3}}{\partial z}\\
&=\frac{\partial f}{\partial x} F_{1}+\frac{\partial f}{\partial y} F_{2}+\frac{\partial f}{\partial z} F_{3}+f\left(\frac{\partial F_{1}}{\partial x}+\frac{\partial F_{2}}{\partial y}+\frac{\partial F_{2}}{\partial z}\right)\\
&=(\nabla f) \cdot \mathbf{F}+f(\nabla \cdot \mathbf{F})
\end{align*}
