 Solution

For a fixed $x\in [0,1]$, the allowed $y$ interval is $[0,\sqrt{x}]$. Integrating the joint probability density over it gives the marginal densityFor a fixed $y\in [0,1]$, the inequality $y\le \sqrt{x}$ is equivalent to $x\ge y^2$. The other marginal density is thereforeAs a normalization check, ${\int }_0^{1}3x^{2}dx=1$ and ${\int }_0^{1}3y(1-y^4)dy=1$.