Answer
$\displaystyle \color{blue}{f_{Y\vert X=x}(y\vert x) = \frac{x+y}{x+\frac{1}{2}},\ 0\le x,y\le 1}$
Work Step by Step
Proceed in a manner similar to that in Example 3.11.5.a-b (see p. 205).
First, obtain the marginal distribution of $X$ using Theorem 3.7.2.
$\begin{align*}
f_X(x) &= \int_{\Omega_Y} f_{X,Y}(x,y)\ dy,\ 0\le x,y\le 1 \\
&= \int_0^1 (x+y)\ dy,\ 0\le x\le 1 \\
&= \left( xy + \frac{y^2}{2}\ \right\vert_{y=0}^{y=1} \\
&= \left( (x)(1) + \frac{1^2}{2}\right) -\left( (x)(0) + \frac{0^2}{2}\right) \\
&= x + \frac{1}{2} - 0 - 0 \\
f_X(x) &= x + \frac{1}{2},\ 0\le x\le 1
\end{align*}$
Next, as in Example 3.11.5.b, use the continuous version of Definition 3.11.1 (see p. 204) to find the desired conditional distribution.
$\begin{align*}
f_{Y\vert X=x}(y\vert x) &= \frac{f_{X,Y}(x,y)}{f_X(x)} \\
&= \frac{(x+y)}{(x+\frac{1}{2})},\ \overbrace{\{(x,y)\vert 0\le x,y\le 1 \}}^{\text{for}\ f_{X,Y}(x,y)} \cap \overbrace{\{x\vert 0\le x\le 1,\ y\in \mathbb{R}\}}^{\text{for}\ f_X(x)} \\
\color{blue}{f_{Y\vert X=x}(y\vert x)}\ &\color{blue}{= \frac{x+y}{x+\frac{1}{2}},\ 0\le x,y\le 1.}
\end{align*}$
