Answer
$\iint_S f \cdot n dS=\iiint_E (\nabla f) dV$
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.
$div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$
$\iint_S fc \cdot n dS=\iiint_Ediv (fc) dV$
or, $\iint_S fc \cdot n dS=\iiint_E f( \nabla \cdot c) +(\nabla f) \cdot c dV$
or, $\iint_S fc \cdot n dS=\iiint_E f(0) +(\nabla f) \cdot c dV$
or, $\iint_S fn \cdot c dS=\iiint_E (\nabla f) \cdot c dV$
and $\iint_S f \cdot n dS=\iiint_E (\nabla f) dV$
Hence the result has been verified.
