Find $g(x|y=\frac{1}{2})$, the conditional pdf of $X$ given $Y = \frac{1}{2}$ (Need confirmation)

Question

Let X and Y be continuous random variables having the joint pdf $$f(x,y) = 8xy , 0\leq{y}\leq{x}\leq{1}$$ I found that the marginal pdf of Y is $f_2(y) = 4y - 4y^3$. Does $g(x|y=\frac{1}{2}) = \frac{f(x,\frac{1}{2})}{f_2(\frac{1}{2})} = \frac{8x}{3}$ or $g(x|y=\frac{1}{2}) = \frac{f(x,y)}{f_2(\frac{1}{2})} = \frac{16xy}{3}$. A little confused here with the definition. Sorry for the tiny fractions, I don't know how to enlarge them.

Answer 1

$$f_Y(y) = \int_{y}^{1} 8xy \: \text{dx} = 4y - 4y^3$$

$$ f_Y(y) = \begin{cases} 4y - 4y^3, & 0 \leq y \leq 1 \\ 0, & \text{otherwise} \end{cases} $$

$$g(x \mid y=\frac{1}{2}) = \frac{f_{X,Y}(x,y=\frac{1}{2})}{f_Y(y=\frac{1}{2})} = \frac{8x}{3}$$