It looks like the boundaries of [tex]R[/tex] are the lines [tex]y=\dfrac23x [/tex] and [tex]y=3x[/tex], as well as the hyperbolas [tex]xy=\frac23[/tex] and [tex]xy=3[/tex]. Naturally, the domain of integration is the set
[tex]R = \left\{(x,y) ~:~ \dfrac{2x}3 \le y \le 3x \text{ and } \dfrac23 \le xy \le 3 \right\}[/tex]
By substituting [tex]x=\frac uv[/tex] and [tex]y=v[/tex], so [tex]xy=u[/tex], we have
[tex]\dfrac23 \le xy \le 3 \implies \dfrac23 \le u \le 3[/tex]
and
[tex]\dfrac{2x}3 \le y \le 3x \implies \dfrac{2u}{3v} \le v \le \dfrac{3u}v \implies \dfrac{2u}3 \le v^2 \le 3u \implies \sqrt{\dfrac{2u}3} \le v \le \sqrt{3u}[/tex]
so that
[tex]R = \left\{(u,v) ~:~ \dfrac23 \le u \le 3 \text{ and } \sqrt{\dfrac{2u}3 \le v \le \sqrt{3u}\right\}[/tex]
Compute the Jacobian for this transformation and its determinant.
[tex]J = \begin{bmatrix}x_u & x_v \\ y_u & y_v\end{bmatrix} = \begin{bmatrix}\dfrac1v & -\dfrac u{v^2} \\\\ 0 & 1 \end{bmatrix} \implies \det(J) = \dfrac1v[/tex]
Then the area element under this change of variables is
[tex]dA = dx\,dy = \dfrac{du\,dv}v[/tex]
and the integral transforms to
[tex]\displaystyle \iint_R 9xy \, dA = \int_{2/3}^3 \int_{\sqrt{2u/3}}^{\sqrt{3u}} \frac{dv\,du}v[/tex]
Now compute it.
[tex]\displaystyle \iint_R 9xy \, dA = \int_{2/3}^3 \ln|v|\bigg|_{v=\sqrt{2u/3}}^{v=\sqrt{3u}} \,du \\\\ ~~~~~~~~ = \int_{2/3}^3 \ln\left(\sqrt{3u}\right) - \ln\left(\sqrt{\frac{2u}3}\right) \, du \\\\ ~~~~~~~~ = \frac12 \int_{2/3}^3 \ln(3u) - \ln\left(\frac{2u}3\right) \, du \\\\ ~~~~~~~~ = \frac12 \int_{2/3}^3 \ln\left(\frac{3u}{\frac{2u}3}\right) \, du \\\\ ~~~~~~~~ = \frac12 \ln\left(\frac92\right) \int_{2/3}^3 du \\\\ ~~~~~~~~ = \frac12 \ln\left(\frac92\right) \left(3-\frac23\right) = \boxed{\frac76 \ln\left(\frac92\right)}[/tex]
