Answer
$\color{blue}{f_Y(y) = \frac{2}{3}(y+1),\ 0\le y\le 1}$
Work Step by Step
Using the continuous version Def. 3.11.1 (see p. 204), we have
$\begin{align*}
f_{Y\vert x}(y) &= \frac{2y+4x}{1+4x},\ 0\le x,y\le 1 \\
\frac{f_{X,Y}(x,y)}{f_X(x)} &= \frac{2y+4x}{1+4x} \\
f_{X,Y}(x,y) &= f_X(x)\cdot \frac{2y+4x}{1+4x} \\
&= \overbrace{\frac{1}{3}\cdot(1+4x)}^{f_X(x)\ \text{as given}} \cdot \frac{2y+4x}{1+4x} \\
f_{X,Y}(x,y) &= \frac{1}{3}\cdot (2y+4x),\ 0\le x,y\le 1
\end{align*}$
We now obtain the marginal distribution of $Y$ by using Theorem 3.7.2 (see p. 167).
$\begin{align*}
f_Y(y) &= \int_{\Omega_X} f_{X,Y}(x,y)\ dx, \ 0\le x,y\le 1 \\
&= \int_0^1 \frac{1}{3}(2y+4x)\ dx \\
&= \frac{1}{3}\left( 2xy+ 4\frac{x^2}{2}\ \right\vert_{x=0}^{x=1} \\
&= \frac{1}{3}\biggl(\left( 2(1)y + 4\frac{1^2}{2} \right) - \left( 2(0)y + 4\frac{0^2}{2} \right) \biggr) \\
&= \frac{1}{3}(2y + 2 - 0 - 0) \\
\color{blue}{f_Y(y)}\ &\color{blue}{= \frac{2}{3}(y+1),\ 0\le y\le 1.}
\end{align*}$
