Answer
$\iint_D |\nabla f|^2 dA=0$
Work Step by Step
$\iint_Df \nabla^2 g dA=\oint_C f(\nabla g) \cdot n ds-\iint_D \nabla f \cdot \nabla g dA$
We need to replace $f$ by $g$.
$\iint_D f \nabla^2 f dA=\oint_C f(\nabla f) \cdot n ds-\iint_D \nabla f \cdot \nabla f dA$
Now, consider the condition when $f$ is harmonic then $\nabla^2 f=0$ on $D$.
$\iint_D f \nabla^2 f dA=\oint_C f(\nabla f) \cdot n ds-\iint_D \nabla f \cdot \nabla f dA \implies \iint_Df \nabla^2 f dA=0$
Further, when $f(x,y)=0$ on the curve $C$.
we get $\iint_Cf( \nabla f) \cdot n ds=0$
This gives: $\iint_D \nabla f \cdot \nabla f dA=0$
Hence, the result $\iint_D |\nabla f|^2 dA=0$
