Caveat: This solution came from playing around with this distribution trying to decompose it into a function of two independent random variables. I think it's neat, but certainly not the standard way to approach things.
Observation: With a little bit of work (basically just change of variables), you can show that if you set $A=X^2,B=X^2/Y^2$, then $A,B$ are independent $U[0,1]$ random variables. Equivalently, if you start with such $A,B$ and set $(X,Y)=\left(\sqrt{A},\sqrt{\frac{A}{B}}\right)$, this will follow the same distribution as the initial distribution given in the question.
Then:
$$\mathbb{P}(Y\le y)=\mathbb{P}\left(\sqrt{\frac{A}{B}}\le y\right)=\mathbb{P}(A\le By^2)$$
$$=\mathbb{E}_B\mathbb{P}(A\le By^2|B)=\mathbb{E}_B(By^2\wedge1)$$
$$=\int_0^1(by^2\wedge1)\,db=y^2\int_0^1(b\wedge y^{-2})\,db$$
$$=\begin{cases}
y^2\int_0^1b\,db=\frac{1}{2}y^2\text{ for } y<1\\
y^2\left[\int_0^{y^{-2}}b\,db+y^{-2}\left(1-y^{-2}\right)\right]=y^2\left[\frac{1}{2y^4}+y^{-2}-y^{-4}\right]=1-\frac{1}{2y^2} \text{ for } y>1
\end{cases}$$
Differentation yields the sought density function.