Answer
We will utilise Jacobians with partitioning to solve for both the parts.
For the first part, consider the following two inverses on two disjoint paritions
\[
\begin{aligned}
X &= \begin{cases}
-\sqrt{Y} &\mbox{$-1 < x < 0$}\newline
\sqrt{Y} &\mbox{$0 \leq x < 1$}
\end{cases}\newline
\implies J &= \begin{cases}
\lvert-\frac{1}{2\sqrt{Y}}\rvert &\mbox{$-1 < x < 0$}\newline
\lvert\frac{1}{2\sqrt{Y}}\rvert &\mbox{$0 \leq x < 1$}
\end{cases} = \frac{1}{2\sqrt{Y}} \quad \text{throughout}\newline
\implies f_{Y}(y) &= \frac{1}{2} \frac{1}{2\sqrt{y}} \quad \forall x
\end{aligned}
\]
However, we must note that the range of \(y\) is same in both the partitions. Hence, we will need to sum up the two pmf to obtain the final pmf of y.
\[
\begin{aligned}
f_{Y}(y) = \frac{1}{2} \frac{1}{2\sqrt{y}} + \frac{1}{2} \frac{1}{2\sqrt{y}} = \frac{1}{2\sqrt{y}} \quad 0 < y < 1
\end{aligned}
\]
For the second part, we partition in a similar manner
\[
\begin{aligned}
X &= \begin{cases}
-\sqrt{Y} &\mbox{$-1 < x < 0$}\newline
\sqrt{Y} &\mbox{$0 \leq x < 3$}
\end{cases}\newline
\implies J &= \begin{cases}
\lvert-\frac{1}{2\sqrt{Y}}\rvert &\mbox{$-1 < x < 0$}\newline
\lvert\frac{1}{2\sqrt{Y}}\rvert &\mbox{$0 \leq x < 3$}
\end{cases} = \frac{1}{2\sqrt{Y}} \quad \text{throughout}\newline
\implies f_{Y}(y) &= \frac{1}{4} \frac{1}{2\sqrt{y}} \quad \forall x
\end{aligned}
\]
Now, the range of \(Y\) is different in the two partitions. Consider the following
\[
\begin{aligned}
f_{Y}(y) &= \begin{cases}
\frac{1}{4} \frac{1}{2\sqrt{y}} &\mbox{$-1 < x < 0$ or $0 < y < 1$}\newline
\frac{1}{4} \frac{1}{2\sqrt{y}} &\mbox{$0 \leq x < 1$ or $0 \leq y < 1$}\newline
\frac{1}{4} \frac{1}{2\sqrt{y}} &\mbox{$1 \leq x < 3$ or $1 \leq y < 9$}
\end{cases}
\end{aligned}
\]
Wherever the partitions overlap for the values of \(y\), we will add the pmf
\[
\begin{aligned}
f_{Y}(y) &= \begin{cases}
\frac{1}{2} \frac{1}{2\sqrt{y}} &\mbox{$0 < y < 1$}\newline
\frac{1}{4} \frac{1}{2\sqrt{y}} &\mbox{$1 \leq y < 9$}
\end{cases}\newline
\end{aligned}
\]
One can verify that the pmf indeed integrates to 1 over the entire support of y \(0 < y < 9\). Another interesting observation is that the total probability over the two regions is \(1/2\) since they divide the original region of \(x\) into two parts as well.