= Solution
For a continuous <random variable> with <probability density function> $f_X$, its <cumulative distribution function> is
$$
F_X(x)=\Pr(X\leq x)=\int_{-\infty}^{x}f_X(u)\,du.
$$
Consequently $F_X'=f_X$ wherever the density is continuous, and almost everywhere for an absolutely continuous distribution.
Here $Y=X^2\geq0$. Thus $F_Y(y)=0$ for $y<0$. For $y\geq0$, the event $X^2\leq y$ is $-\sqrt y\leq X\leq\sqrt y$. Symmetry of the <Laplace distribution> gives
$$
F_Y(y)=\int_{-\sqrt y}^{\sqrt y}\frac12e^{-|x|}\,dx
=\int_0^{\sqrt y}e^{-x}\,dx=1-e^{-\sqrt y}.
$$
Therefore the <square of a standard Laplace variable> has <cumulative distribution function>
$$
\boxed{F_Y(y)=\begin{cases}0,&y<0,\\1-e^{-\sqrt y},&y\geq0.\end{cases}}
$$
It is continuous at zero, so there is no point mass there. As a check, its <probability density function> for $y>0$ is $e^{-\sqrt y}/(2\sqrt y)$, whose integral becomes $\int_0^\infty e^{-u}\,du=1$ after setting $u=\sqrt y$.
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