= Solution
In the positive quadrant, the inverse transformation is single-valued:
$$
x=\sqrt{v/u},\qquad y=\sqrt{uv}.
$$
The two rays become $1\le u\le2$ and the two <hyperbolas> become $1\le v\le3$, so the curved original region becomes a rectangle. These are the middle two panels of the figure in part 2i. The forward <Jacobian determinant> is
$$
\frac{\partial(u,v)}{\partial(x,y)}
=\begin{vmatrix}-y/x^2&1/x\\y&x\end{vmatrix}=-2u,
\qquad dx\,dy=\frac{du\,dv}{2u}.
$$
Since $x^3y=v^2/u$, the required <double integral> is
$$
\int_1^2\int_1^3\frac{v^2}{2u^2}\,dv\,du
=\frac12\left(\frac12\right)\left(\frac{26}{3}\right)
=\boxed{\frac{13}{6}.}
$$
The signs of $x,y$ and $u>0$ justify the chosen square-root branch and the simplified absolute <Jacobian determinant>.
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