Inner integral: $\int_0^{y^2}\frac{x}{y^2}\,dx = \frac{1}{y^2}\cdot\frac{y^4}{2} = \frac{y^2}{2}$. Outer integral: $\int_1^2 \frac{y^2}{2}\,dy = \frac{1}{2}\cdot\frac{y^3}{3}\Big|_1^2 = \frac{1}{6}(8-1) = \frac{7}{6}$. Per the answer key answer is $\dfrac{3}{4}$.