Answer
$$1.028$$
Work Step by Step
\begin{aligned}
\int_{1}^{2} \int_{1}^{3} \frac{\ln (x y) d y d x}{y} &=\int_{1}^{2}\left(\left.\frac{1}{2}[\ln (x y)]^{2}\right|_{1} ^{3 }\right) d x \\
&=\frac{1}{2} \int_{1}^{2}[\ln (3 x)]^{2}-[\ln (x)]^{2} d x \\
&=\frac{1}{2} \int_{1}^{2}[\ln (3 x)]^{2} d x-\frac{1}{2} \int_{1}^{2}[\ln (x)]^{2} d x \\
&=\frac{1}{2}\left[x(\ln 3 x)^{2}-(\ln 3)^{2}\right]-\int_{1}^{2} \ln (3 x) d x-\frac{1}{2}\left[2(\ln 2)^{2}-0\right]+\int_{1}^{2} \ln x d x \\
&=(\ln 6)^{2}-\frac{1}{2}(\ln 3)^{2}-\left[x \ln (3 x)-\left.x\right|_{1} ^{2 }\right]-(\ln 2)^{2}+\left[x \ln x-\left.x\right|_{1} ^{2 }\right]\\
&= (\ln 6)^{2}-\frac{1}{2}(\ln 3)^{2}-(\ln 2)^{2}-2 \ln 6+\ln 3+2 \ln 2 \approx 1.028
\end{aligned}
