= Solution
Complete the square in the transformation:
$$
Y=X(2-X)=1-(X-1)^2.
$$
Since $0<X<2$ almost surely, $0<Y\leq1$. For $0\leq y<1$, put $r=\sqrt{1-y}$. The <event> $Y>y$ is equivalent to $1-r<X<1+r$. Integrating the given <probability density function> over both branches at once gives the <survival function>
$$
\begin{aligned}
\mathbb P(Y>y)&=\int_{1-r}^{1+r}\frac x2\,dx
=\frac{(1+r)^2-(1-r)^2}{4}=r.
\end{aligned}
$$
Thus, including the tails,
$$
\boxed{\mathbb P(Y>y)=\begin{cases}1&y<0,\\\sqrt{1-y}&0\leq y<1,\\0&y\geq1.\end{cases}}
$$
The <cumulative distribution function> on $(0,1)$ is $1-\sqrt{1-y}$. Differentiation gives
$$
\boxed{f_Y(y)=\begin{cases}\dfrac1{2\sqrt{1-y}}&0<y<1,\\0&\text{otherwise}.\end{cases}}
$$
Both inverse branches matter: $x=1\pm\sqrt{1-y}$. The endpoint divergence is integrable, and $\mathbb P(Y=1)=\mathbb P(X=1)=0$, so no endpoint atom is needed.
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