[tex]X[/tex] has CDF
[tex]F_X(x)=\mathbb P(X\le x)=\begin{cases}1-e^{-\lambda x/2}&\text{for }x\ge0\\0&\text{otherwise}\end{cases}[/tex]
The CDF of [tex]Y[/tex] is then
[tex]F_Y(y)=\mathbb P(Y\le y)=\mathbb P(X^2\le y)=\mathbb P(X\le\sqrt y)=F_X(\sqrt y)[/tex]
[tex]\implies F_Y(y)=\begin{cases}1-e^{-\lambda\sqrt y/2}&\text{for }y\ge0\\0&\text{otherwise}\end{cases}[/tex]
[tex]\implies F_Y(y)=\begin{cases}1-e^{-(y/(4/\lambda^2))^{1/2}}&\text{for }y\ge0\\0&\text{otherwise}\end{cases}[/tex]
which is the CDF of a Weibull distribution with shape parameter [tex]\dfrac4{\lambda^2}[/tex] and scale parameter [tex]\dfrac12[/tex].
