= Solution
{c}
The support condition $-y<x<y$ forces $y>0$. As a check, the <joint probability density> integrates to
$$
\int_0^\infty\int_{-y}^{y}2e^{-2y}\,dx\,dy=\int_0^\infty4ye^{-2y}\,dy=1.
$$
For any real $x$, the <marginal density> of $X$ is
$$
f_X(x)=\int_{|x|}^\infty2e^{-2y}\,dy=e^{-2|x|}>0.
$$
Dividing the <joint probability density> by this <marginal density> gives the <conditional density>:
$$
\boxed{f_{Y\mid X=x}(y)=2e^{-2(y-|x|)}\mathbf1_{\{y>|x|\}}.}
$$
Thus, conditionally on $X=x$, $Y-|x|$ has the rate-two <exponential distribution>.
For $U=X^2$, use the two inverse branches $x=\sqrt u$ and $x=-\sqrt u$. Each branch has absolute <Jacobian determinant> $1/(2\sqrt u)$. Both lie in the original support exactly when $u>0$ and $y>\sqrt u$. The <change of variables for probability densities> therefore yields
$$
\boxed{f_{U,Y}(u,y)=\begin{cases}\displaystyle\frac{2e^{-2y}}{\sqrt u},&u>0,\ y>\sqrt u,\\0,&\text{otherwise}.\end{cases}}
$$
No <atom of a measure> is created at $u=0$, since $\mathbb P(X=0)=0$. The value of the <probability density function> at this boundary may be assigned arbitrarily. Checking the <integral> over $0<u<y^2$ gives $4ye^{-2y}$, which integrates to one.
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