So the question asks:
Let $X, Y$ be random variables, with joint probability density function:
$f_{X,Y}(x,y) = \left\{ \begin{array}{ 1 l } 0.25ye^{-y} & \mbox{if $0≤|x|≤y$}\\ 0 & \mbox{ otherwise} \end{array} \right.$
Find the conditional density $f_{X|Y =y} (x)$ and identify the conditional distribution $X|_{Y =y}$. Use the law of total expectation to find $E[X^2]$.
So so far I have:
$f_Y(y) = \int_{ -y }^{y} 0.25ye^{-y} \,dx $ =$0.5e^{-y}y^2$
$f _{X|Y=y} (x) =f(x,y)/f_Y (y) = 0.25ye^{-y}/0.5e^{-y} y^2 = 0.5/y$
normal distribution
But I am a little confused about the law of total expectation
So does $E(x^2) = E[E(X^2|Y)]$ = $\sum_{0}^\infty E[X^2|Y]P[Y] $
But how do I find the probability of Y and the expectation?
Yes, $$E[X^2] = E[E[X^2 \mid Y]]$$ but to make your life easier, you should start by finding $E[X^2 \mid Y]$ first. It's not as difficult as you might think: just remember $$E[X] = \int_{-\infty}^{\infty}xf_{X}(x)\text{ d}x$$ so analogously, $$E[X^2 \mid Y] = \int_{-\infty}^{\infty}x^2f_{X \mid Y}(x \mid y)\text{ d}x\text{.}$$ Then after integrating over $X$, you integrate over $Y$ (which is the "random portion" left), so $$E[E[X^2 \mid Y]] = \int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}x^2f_{X \mid Y}(x \mid y)\text{ d}x\right)\text{ d}y\text{.}$$ $f_{X \mid Y}(x \mid y)$ is the same as $f_{X \mid Y = y}(x)$ in your notation.