Answer
$\color{blue}{f_Y(y) = \dfrac{y}{6} + \dfrac{y^2}{4},\ 0\lt y\lt 2}$
Work Step by Step
Taking the first derivative of $F_Y(y)$ with respect to $y$ where it is differentiable gives:
$\begin{align*}
F_Y(y) &= {\small \frac{1}{12}}(y^2+y^3),\ 0\le y\le 2 \\
\dfrac{dF_Y(y)}{dy} &= {\small \frac{1}{12}}(2y + 3y^2),\ 0\lt y\lt 2 \\
&= \frac{2}{12}y + \frac{3}{12}y^2,\ 0\lt y\lt 2 \\
\color{blue}{f_Y(y)} &\color{blue}{= \frac{y}{6} + \frac{y^2}{4},\ 0\lt y\lt 2}
\end{align*}$
