I solved this integral
$$\int_0^{ln2} \int_{e^y}^2 \frac xydxdy$$
and got this: $$\frac{-\ln(2)^2}8 -\frac{\ln(2)}8 + \frac{3}{16} $$
However, when I checked the answer from the back of the book, the answer appears as
$$\frac{\ln^3(2)}6 $$
Probably, it's very apparent to many people but I couldn't simply down to that. I'd appreciate it if anyone can show me the steps of simplification.
The answer in the book is correct, and has approximate value $$ \frac{\log^3 2}{6} \approx 0.055504108664821579953.$$ Your value is approximately $0.040799975690231658239$, too small. Indeed, evaluating the inner integral yields $$\int_{x=e^y}^2 \frac{y}{x} \, dx = \left[y \log x \right]_{x=e^y}^2 = y (\log 2 - y).$$ Evaluating the outer integral gives $$\int_{y=0}^{\log 2} y (\log 2 - y) \, dy = \left[\frac{y^2}{2} \log 2 - \frac{y^3}{3} \right]_{y=0}^{\log 2} = \frac{\log^3 2}{2} - \frac{\log^3 2}{3} = \frac{\log^3 2}{6},$$ as claimed.