Answer
$\color{blue}{F_Y(y) = \begin{cases} 0,& y\lt 0 \\ 1 -e^{-y} -ye^{-y}, & y\ge 0 \end{cases}}$
Work Step by Step
$\begin{align*}
F_Y(y) &= \int_{-\infty}^y f_Y(u)\ du \\
&= \begin{cases} \displaystyle \int_{-\infty}^y 0\ du,& y\lt 0\\ \displaystyle \int_{-\infty}^0 0\ du + \underbrace{\int_0^y ue^{-u}\ du}_{\begin{array}{rl}
\text{Use integration} & \text{by parts} \\
m=u & dn=e^{-u}\ du \\
dm=du & n=-e^{-u}
\end{array}}, &y\ge 0 \end{cases} \\
&= \begin{cases} 0,& y\lt 0 \\ \displaystyle 0 + \left. -ue^{-u}\right]_0^y - \int_0^y (-e^{-u})\ du, & y\ge 0 \end{cases} \\
&= \begin{cases} 0,& y\lt 0 \\ \displaystyle (-ye^{-y} - 0) + \left( -e^{-u}\right]_0^y, & y\ge 0 \end{cases} \\
&= \begin{cases} 0,& y\lt 0 \\ \displaystyle -ye^{-y} + \left( -e^{-y} - (-e^0) \right), & y\ge 0 \end{cases} \\
\color{blue}{F_Y(y)} &\color{blue}{= \begin{cases} 0,& y\lt 0 \\ 1 -e^{-y} -ye^{-y}, & y\ge 0 \end{cases}} \\
\end{align*}$
