= Solution
The inequalities $y\le x\le2y$ force $y\ge0$, and the reciprocal bounds exclude zero. Thus the whole region lies in the positive quadrant. Introduce <ratio-product coordinates>
$$
u=\frac{x}{y},\qquad v=xy.
$$
The region becomes the rectangle $1\le u\le2$, $1\le v\le2$. The inverse map and its <Jacobian determinant> are
$$
x=\sqrt{uv},\qquad y=\sqrt{v/u},\qquad
\frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix}x/(2u)&x/(2v)\\-y/(2u)&y/(2v)\end{vmatrix}=\frac{xy}{2uv}=\frac1{2u}>0.
$$
The <change of variables formula> then gives
$$
\int_A\frac{x}{y}\,dx\,dy=\int_1^2\int_1^2u\frac1{2u}\,dv\,du=\boxed{\frac12}.
$$
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