Answer
$\iint_S (f \nabla g) \cdot n dS=\iiint_E (f \nabla^2g+\nabla f \cdot \nabla g)dV$
Work Step by Step
We know that $D_nf=(\nabla f) \cdot n$
Here, we have $\iint_S (f \nabla g) dS=\iiint_E div (f \nabla g) dV=\iint_E \nabla (F \nabla g) dV$
We have $F=\nabla g$
Then $\iint_S (f \nabla g) dS=\iiint_E div (f \nabla g) dV=\iint_E \nabla (F \nabla g) dV$ becomes:
$\implies \iint_S (f \nabla g) dS=\iiint_E div (f \nabla g) dV$
$\implies \iint_E \nabla (f \nabla g) dV=\iiint_E f(\nabla \cdot ( \nabla g) +\nabla f \cdot (\nabla g) dV$
Thus, we have $\iint_S (f \nabla g) \cdot n dS=\iiint_E (f \nabla^2g+\nabla f \cdot \nabla g)dV$ (Verified)
