Find the conditional variance of y given $x = \frac{1}{2}$
$$ f(x,y)=\left\{ \begin{array}{c} \frac{2}{3}(x+2y) \text{ for } 0<x<1,0<y<1 \\ 0 \qquad \text{elsewhere} \\ \end{array} \right. $$
In order to solve this one must substitute in for $x=\frac{1}{2}$.
Thus
$$f(\frac{1}{2}|y) =\left\{ \begin{array}{c} \frac{1+4y}{3}\qquad \text{ for } 0<x<1,0<y<1 \\ 0 \qquad \text{elsewhere} \\ \end{array} \right. $$
$$\Rightarrow E(\frac{1}{2}|y)= \int^1_0y(\frac{1+4y}{3})dy =\frac{11}{18} $$
$$\Rightarrow E(\frac{1}{2}|y^2)= \int^1_0y^2(\frac{1+4y}{3})dy=\frac{4}{9} $$
$$\Rightarrow \sigma^2_{\frac{1}{2}|y}=\frac{4}{9}-(\frac{11}{18})^2=\frac{23}{324}$$
Would this be the correct way to solve this problem according to what the question has asked for?
You need to normalize $g(y)=f(x,y)|_{x=1/2}$, so that $\int_0^1g(y)dy=1$.
Happily, in this case that integral is already $1$, so the normalizing factor is $1$.