Answer
$\iint_S (f \nabla g) \cdot n dS=\iiint_E (f \nabla^2g+\nabla f \cdot \nabla g)dV$
Work Step by Step
Since, we have $D_nf=(\nabla f) \cdot n$
Divergence Theorem: $\iiint_Ediv \overrightarrow{F}dV=\iint_S \overrightarrow{F}\cdot d\overrightarrow{S} $
where, $div F=\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}$
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, $\iint_S (f \nabla g) dS=\iiint_E div (f \nabla g) dV$
and $\iint_E \nabla (F \nabla g) dV=\iiint_E f(\nabla \cdot ( \nabla g) +\nabla f \cdot (\nabla g) dV$
Thus, $\iint_S (f \nabla g) \cdot n dS=\iiint_E (f \nabla^2g+\nabla f \cdot \nabla g)dV$ (Verified)
