Answer
$\iint_S (f \nabla g) \cdot n dS=\iiint_E (f \nabla^2g+\nabla f \cdot \nabla g)dV$
Work Step by Step
The Divergence Theorem states that $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $
Here, $S$ is a closed surface and $E$ is the region inside that surface.
$div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$
Since, we have $D_nf=(\nabla f) \cdot n$
This implies that $\iint_S (f \nabla g) dS=\iiint_E div (f \nabla g) dV=\iint_E \nabla (F \nabla g) dV$
Since, $F=\nabla g$
Therefore, $\iint_S (f \nabla g) dS=\iiint_E div (f \nabla g) dV=\iint_E \nabla (F \nabla g) dV=\iiint_E f(\nabla \cdot ( \nabla g) +\nabla f \cdot (\nabla g) dV$
Hence, it has been verified that $\iint_S (f \nabla g) \cdot n dS=\iiint_E (f \nabla^2g+\nabla f \cdot \nabla g)dV$
