Answer
$\int_{R} dx dy=\iint_{S} [\dfrac{\partial (x,y)}{\partial (u,v) }]$
Therefore, the result has been proved.
Work Step by Step
Green's Theorem states that:
$\oint_C M \,dx+ N \,dy=\iint_{D}(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y})dA$
We need to work out with the line integral and evaluate the integrand of the double integral as follows:
$A=\oint_{C} x dy$
Since, $\int_{R} dx dy=\oint_{\partial R} x dy$
and $x = g(u,v); y= h(u,v)$
Now, $dy= \dfrac{\partial h}{\partial u} du+ \dfrac{\partial h}{\partial v} dv$
Now, $\oint_{\partial R} x dy=\oint_{\partial S} [g(u,v) \dfrac{\partial h}{\partial u}] \ du+ [g(u,v)\dfrac{\partial h}{\partial v}] \ dv$
Green's Theorem states that:
$\oint_CP\,dx+Q\,dy=\iint_{D}(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y})\ dA =\iint_{\partial S} ( \dfrac{\partial g}{\partial u} \dfrac{\partial h}{\partial v}-\dfrac{\partial g}{\partial v} \dfrac{\partial h}{\partial u})+ g[\dfrac{\partial }{\partial u}(\dfrac{\partial h}{\partial v})-\dfrac{\partial }{\partial v}(\dfrac{\partial h}{\partial u})] dA$
and $\int_{R} dx dy=\iint_{S} [\dfrac{\partial (x,y)}{\partial (u,v) }]$
Therefore, the result has been proved.
