With [tex]Y=X^2[/tex], note that the support of [tex]Y[/tex] will be all non-negative real numbers.
[tex]F_Y(y)=\mathbb P(Y\le y)[/tex]
[tex]F_Y(y)=\mathbb P(X^2\le y)[/tex]
[tex]F_Y(y)=\mathbb P(|X|\le\sqrt y)[/tex]
[tex]F_Y(y)=\mathbb P(-\sqrt y\le X\le\sqrt y)[/tex]
[tex]F_Y(t)=\mathbb P(X\le\sqrt y)-\mathbb P(X\le-\sqrt y)[/tex]
[tex]F_Y(y)=F_X(\sqrt y)-F_X(-\sqrt y)[/tex]
Since [tex]X\sim\mathcal N(0,1)[/tex], you have
[tex]F_X(x)=\mathbb P(X\le x)=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^xe^{-t^2/2}\,\mathrm dt=\frac12+\frac12\mathrm{erf}\left(\dfrac x{\sqrt2}\right)[/tex]
(where [tex]\mathrm{erf}(x)[/tex] denotes the error function) and so
[tex]F_Y(y)=\left(\dfrac12+\dfrac12\mathrm{erf}\left(\sqrt{\dfrac y2}\right)\right)-\left(\dfrac12+\dfrac12\mathrm{erf}\left(-\sqrt{\dfrac y2}\right)\right)[/tex]
[tex]F_Y(y)=\mathrm{erf}\left(\sqrt{\dfrac y2}}\right)[/tex]
[tex]\implies f_Y(y)=\dfrac{\mathrm dF_Y(y)}{\mathrm dy}=\dfrac1{\sqrt{2\pi}}\dfrac{e^{-y/2}}{\sqrt y}[/tex]
where [tex]y\ge0[/tex].
