That is mostly okay, but you are making way too much work for yourself. You are asked to transform variable to find a pdf from a pdf. You don't actually have to perform an integration to a CDF to get there. You can apply the chain rule of differentiation to go straight to goal.
$$\begin{align}U_1&=3Y\\\mathsf P(U_1\leqslant u) &= \mathsf P(3Y\leqslant u)
\\ & = \mathsf P(Y\leqslant u/3)
\\ F_{U_1}(u) & = F_Y(u/3)
\\ f_{U_1}(u) & = \tfrac 13 f_Y(u/3) \\ & = \tfrac 12 (\tfrac u3)^2~\mathbf 1_{-1\leqslant (u/3)\leqslant 1}\\ &=\tfrac 1{18}u^2~\mathbf 1_{(-3\leqslant u\leqslant 3)}\end{align}$$
When the transformation $Y\mapsto h(Y)$ has an inverse function $g()$, then the fact that the pdf is the unsigned derivative of the CDF means:
$$F_{h(Y)}(u)=F_Y(g(u)) \implies f_{h(Y)}(u) = \lvert g'(u)\rvert~f_Y(g(u))$$
Thus we do not have to invoke the CDF at all:
$$\begin{align}U_2 & =3-Y\\ f_{U_2}(u) ~&=f_{3-Y}(u) \\ &= \lvert\tfrac{\partial (3-u)}{\partial u}\rvert f_Y(3-u) \\ &= \tfrac {3}{2} (3-u)^2 ~\mathbf 1_{(-1\leqslant 3-u\leqslant 1)}\\ &=\tfrac {3}{2}(u-3)^2~\mathbf 1_{(2\leqslant u\leqslant 4)}\end{align}$$
However, complications do arise when the transformation $Y\mapsto U$ is not invertable. In this case $Y\mapsto Y^2$ does not have a single inverse over $[-1;1]$. Taking it carefully we find:
$$\begin{align}U_3&=Y^2 \\\mathsf P(U_3\leqslant u) &= \mathsf P(Y^2\leqslant u)
\\ & = \mathsf P({-}\surd u\leqslant Y\leqslant \surd u)
\\ F_{U_3}(u) & = F_Y(\surd u)-F_Y({-}\surd u)
\\ f_{U_3}(u) & = \tfrac 1{2\surd u} f_Y(\surd u)~+~\tfrac 1{2\surd u} f_Y({-}\surd u))
\\ & = \tfrac 3{4\surd u}((\surd u)^2\mathbf 1_{-1\leqslant \surd u\leqslant 1}+({-}\surd u)^2\mathbf 1_{-1\leqslant {-}\surd u\leqslant 1})
\\ & = \tfrac 3 2 \surd u~\mathbf 1_{0\leqslant u\leqslant 1}\end{align}$$
$\blacksquare$
What you did makes a lot of sense, but what if it $v-2$ was a saddle point?
A better way to see it is using the double derivative test, i.e.
\begin{equation}
f'(x) =
\frac{\big(-\frac{1}{2}\big)x^{\frac{v}{2} - 1}e^{-\frac{x}{2}} + ( \frac{v}{2} - 1) x^{\frac{v}{2} - 2}e^{-\frac{x}{2}}} {2^{\frac{v}{2}} \Gamma(v/2)}
\end{equation}
Then
\begin{equation}
f''(x)
=
\frac{1}{2^{\frac{v}{2}} \Gamma(v/2)}
\dfrac{x^{\frac{v}{2}-3}\left(x^2+\left(4-2v\right)x+v^2-6v+8\right)\mathrm{e}^{-\frac{x}{2}}}{4}
\end{equation}
At the extremum, we have
\begin{equation}
f''(v - 2)
=
\frac{(v-2)^{\frac{v}{2}-3}}{4(2^{\frac{v}{2}} \Gamma(v/2))}
( (v-2)^2 -2(v-2)(v-2) + (v-2)(v-4) )e^{-\frac{v-2}{2}}
\end{equation}
that is
\begin{equation}
f''(v - 2)
=
\frac{(v-2)^{\frac{v}{2}-3}e^{-\frac{v-2}{2}}}{4(2^{\frac{v}{2}} \Gamma(v/2))}
( -(v-2)^2 + (v-2)(v-4) )
\end{equation}
which is
\begin{equation}
f''(v - 2)
=
-2
\frac{(v-2)^{\frac{v}{2}-3}e^{-\frac{v-2}{2}}}{4(2^{\frac{v}{2}} \Gamma(v/2))}
(v-2)
\end{equation}
For $v > 2$ it is easy to see that $f''(v-2) < 0$, hence it is a maximum.
Best Answer
This part is wrong. This holds only when the change of variables is invertible. And your function $x \mapsto \sin x$ is definitely not invertible on $(0,2\pi)$.
In this case you may want to get into the basic principles. The p.d.f. of a continuous random variable is defined as the derivative of the c.d.f. of $X$
$$ f_X(x) = \frac{d}{dx}\mathbb{P}(X \leq x) $$
and we know how to deal with c.d.f. In our case, $Y$ can take values in $[-1,1]$ so it is only meaningful to investigate the range $y \in [-1, 1]$. Then
$$\mathbb{P}(Y \leq y) = \mathbb{P}(\sin X \leq y) = \frac{2\pi - 2\arccos y}{2\pi}. $$
(If you can't see this easily, consider two cases $y \in [0, 1]$ and $y \in [-1, 0]$ separately and try to read out the above probability from the graph of the sine function.) So we have
$$ f_Y(y) = \frac{d}{dy}\mathbb{P}(Y \leq y) = \frac{1}{\pi\sqrt{1-y^2}} $$
when $y \in [-1, 1]$ and $f_Y(y) = 0$ outside $[-1, 1]$.