Answer
$\color{blue}{f_W(w) = \dfrac{(w-1)^3}{4},\ 1\lt w \lt 3}$.
$\color{blue}{f_W(w) \ne 0,\ 1\lt w\lt 3}$.
Work Step by Step
$\underline{\text{Find the cdf of}\ W}$
$\begin{align*}
F_W(w) &= P(W\le w) \\
&= P(2Y+1\le w) \\
&= P(Y\le (w-1)/2) \\
&= \int_{-\infty}^{(w-1)/2} f_Y(y)\ dy \\
&= \begin{cases} \displaystyle \int_{-\infty}^0 f_Y(y)\ dy, & (w-1)/2\lt 0 \\ \displaystyle \int_{-\infty}^0 f_Y(y)\ dy + \int_0^{(w-1)/2} f_Y(y)\ dy, & 0\le (w-1)/2 \lt 1 \\ \displaystyle \int_{-\infty}^0 f_Y(y)\ dy + \int_0^1 f_Y(y)\ dy + \int_1^{(w-1)/2}, & (w-1)/2 \ge 1 \end{cases} \\
&= \begin{cases} \displaystyle \int_{-\infty}^0 0\ dy, & w\lt 1 \\ \displaystyle \int_{-\infty}^0 0\ dy + \int_0^{(w-1)/2} 4y^3\ dy, & 1\le w \lt 3 \\ \displaystyle \int_{-\infty}^0 0\ dy + \int_0^1 4y^3\ dy + \int_1^{(w-1)/2} 0\ dy, & w \ge 3 \end{cases} \\
&= \begin{cases} 0 & w\lt 1 \\ 0 + \left( y^4\right]_0^{(w-1)/2} , & 1\le w \lt 3 \\ \displaystyle 0 + \left( y^4\right]_0^1 + 0, & w \ge 3 \end{cases} \\
&= \begin{cases} 0 & w\lt 1 \\ \left[\left(\frac{w-1}{2}\right)^4 -0^4\right] , & 1\le w \lt 3 \\ \left(1^4 -0^4\right), & w \ge 3 \end{cases} \\
F_W(w) &= \begin{cases} 0 & w\lt 1 \\ \left(\frac{w-1}{2}\right)^4 , & 1\le w \lt 3 \\ 1, & w \ge 3 \end{cases}
\end{align*}$
It follows that
$\begin{align*}
\dfrac{F_W(w)}{dw} &= \begin{cases} 0 & w\lt 1 \\ 4\left(\frac{w-1}{2}\right)^3\left(\frac{1}{2}\right) , & 1\le w \lt 3 \\ 0, & w \gt 3 \end{cases} \\
f_W(w) &= \begin{cases} 0 & w\lt 1 \\ \frac{(w-1)^3}{4}, & 1\le w \lt 3 \\ 0, & w \gt 3 \end{cases}
\end{align*}$
Or simply, $\color{blue}{f_W(w) = \dfrac{(w-1)^3}{4},\ 1\lt w \lt 3},$ since $\color{blue}{f_W(w) \ne 0,\ 1\lt w\lt 3}$, and $0$ otherwise.
