Answer
$\int_Cf_y dx-f_x dy$ is independent of the path in any simple region $D$
Work Step by Step
Consider the Green's Theorem: $\int_C fdx+g dy=\iint_D (\dfrac{\partial g}{\partial x}-\dfrac{\partial f}{\partial y})dA$
Here, $D$ is the region enclosed inside the counter-clockwise oriented loop $C$.
Consider $\nabla^2 f=0$ and $\nabla(\nabla f)=0$
and $\dfrac{\partial^2 f}{\partial x^2}+\dfrac{\partial^2 f}{\partial y^2}=0$
Now, we have $\int_Cf_y dx+(-f_x) dy=\iint_D (\dfrac{\partial f_x}{\partial x}-\dfrac{\partial (-f_y)}{\partial y})dA$
or, $=\iint_D(\dfrac{\partial^2 f}{\partial x^2}+\dfrac{\partial^2 f}{\partial y^2})dA $
or, $\int_Cf_y dx-f_x dy=\iint_D (0)dA=0$
It has been proved that the integral $\int_Cf_y dx-f_x dy$ is independent of the path in any simple region $D$
