Answer
$\iint_D div (f \nabla g) dA=\oint_C f(\nabla g) \cdot n ds$
and $\iint_Df \nabla^2 g dA=\oint_C f(\nabla g) \cdot n ds-\iint_D \nabla f \cdot \nabla g dA$
Work Step by Step
$\iint_Df \nabla^2 g dA+\iint_D \nabla f \cdot \nabla g dA=\oint_C f(\nabla g) \cdot n ds-\iint_D \nabla f \cdot \nabla g dA+\iint_D \nabla f \cdot \nabla g dA$
This implies that
$\iint_Df \nabla^2 g dA+\iint_D \nabla f \cdot \nabla g dA=\oint_C f(\nabla g) \cdot n ds$
Re-arrange as: $\iint_D \nabla (f \nabla g) dA=\oint_C f(\nabla g) \cdot n ds$
This implies that $f\nabla^2 g+\nabla f \cdot \nabla g=f\nabla \cdot (\nabla g)+\nabla f \cdot \nabla g=\nabla \cdot (f \nabla g)$
$\iint_D div (f \nabla g) dA=\oint_C f(\nabla g) \cdot n ds$
Hence, the result has been proved.
$\iint_Df \nabla^2 g dA=\oint_C f(\nabla g) \cdot n ds-\iint_D \nabla f \cdot \nabla g dA$
