Answer
$\iint_Df \nabla^2 g dA=\oint_C f(\nabla g) \cdot n ds-\iint_D \nabla f \cdot \nabla g dA$
Work Step by Step
We need to prove that $\iint_Df \nabla^2 g dA=\oint_C f(\nabla g) \cdot n ds-\iint_D \nabla f \cdot \nabla g dA$
$\iint_Df \nabla^2 g dA+\iint_D \nabla f \cdot \nabla g dA=\oint_C f(\nabla g) \cdot n ds-\iint_D \nabla f \cdot \nabla g dA+\iint_D \nabla f \cdot \nabla g dA$
or, $\iint_Df \nabla^2 g dA+\iint_D \nabla f \cdot \nabla g dA=\oint_C f(\nabla g) \cdot n ds$
or, $\iint_D \nabla (f \nabla g) dA=\oint_C f(\nabla g) \cdot n ds$
or, $\iint_D div (f \nabla g) dA=\oint_C f(\nabla g) \cdot n ds$
Hence, $\iint_Df \nabla^2 g dA=\oint_C f(\nabla g) \cdot n ds-\iint_D \nabla f \cdot \nabla g dA$
