I had to compute the surface area of portion of surface $z^2 = 2xy$ which lies above the first quadrant of X-Y planeand is cut off by the planes $x=2$ and $y=1$. Here's my work :
Since,
$$A(S) = \int \int_{D} \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \ dA$$
First I took $z = \sqrt{2xy}$ and then I took $D$ as $0 \leq y \leq 1$ and $0 \leq x \leq 2$ and doubled the result I got.
Is my method correct?
Your method looks fine, except "above the first quadrant" usually denotes $z\geq 0$ so doubling the result would be unnecessary. So you should get:
$$\displaystyle A(S)=\int_{0}^{1}\int_{0}^{2}dx\,dy\sqrt{1+\frac{y}{2x}+\frac{x}{2y}}=4$$