Answer
$div (\nabla f \times \nabla g)=0$
Work Step by Step
Here, we have $div (F \times G)=\nabla \cdot (F \times G)$
$div (F \times G)=\dfrac{\partial}{\partial x}[G_3F_2-G_2F_3]-\dfrac{\partial}{\partial y}[G_3F_1-G_1F_3]+\dfrac{\partial}{\partial z}[G_2F_1-G_1F_2]$
or, $=[G_1(\dfrac{\partial F_3}{\partial y}-\dfrac{\partial F_2}{\partial z})-G_2(\dfrac{\partial F_3}{\partial x}-\dfrac{\partial F_1}{\partial z})+G_3(\dfrac{\partial F_2}{\partial x}-\dfrac{\partial F_1}{\partial y})]-[F_1(\dfrac{\partial G_3}{\partial y}-\dfrac{\partial G_2}{\partial z})-F_2(\dfrac{\partial G_3}{\partial x}-\dfrac{\partial G_1}{\partial z})+F_3(\dfrac{\partial G_2}{\partial x}-\dfrac{\partial G_1}{\partial y})]$
or, $=(curl F) \cdot G-F \cdot (curl G)$
Further, $div (\nabla f \times \nabla g)=(curl (\nabla f)) \cdot (\nabla g)-(\nabla f) \cdot (curl (\nabla g))$
or, $=(0) \cdot (\nabla g)-(\nabla f) \cdot (0)=0$
Hence, we get $div (\nabla f \times \nabla g)=0$ (proved)
