Answer
$\oint_C D_n g ds=0$
Work Step by Step
Since, $D_ng$ is defined as $\nabla g \cdot n$
$\oint_C F \cdot n ds=\iint_D div F(x,y) dA$
when $\nabla^2 g=0$, that is, $\oint_C \nabla g \cdot n ds=0$
and when $F=\nabla g$
$\oint_C \nabla g \cdot n ds=\iint_D div (\nabla g) dA=\iint_D \nabla \cdot (\nabla g) dA=\iint_D \nabla^2 g dA=\iint_D (0) dA=0$
This yields $\oint_C D_n g ds=0$
Hence, the result is proved.
