confusion regarding pdf? I am reading a question regarding probability
I have attached snaps of both question and answer
I am confused regarding encircled equations of answer especially how we proceed from 2nd row/equation to third ,more specifically we are using a substitution of $y^1/2$ in place of x,but then we should see $e^-(y^1/2) /2$,but we see  $e^-(y) /2$
Below is question

Below is answer

 A: It should be $$f_X(x)=\dfrac{\mathrm e^{-x^2/2}}{\surd(2\pi)}~\mathbf 1_{x\in\Bbb R}$$
Then because we have $$x_1(y)=\surd y~\mathbf 1_{y\in[0..\infty)}\\x_2(y)=-\surd y~\mathbf 1_{y\in(0..\infty)}$$, ...
$$\begin{align}
f_Y(y) &= f_X(x_1(y))\cdot\lvert x_1'(y)\rvert + f_X(x_2(y))\cdot\lvert x_2'(y)\rvert
\\[1ex]&= f_X(\surd y)\cdot\left\lvert \dfrac{\mathrm d (y^{1/2})}{\mathrm d y}\right\rvert~\mathbf 1_{y\in[0..\infty)}+f_X(-\surd y)\cdot\left\lvert \dfrac{\mathrm d (-y^{1/2})}{\mathrm d y}\right\rvert~\mathbf 1_{y\in(0..\infty)}
\\[1ex]&= \dfrac{\mathrm e^{-y/2}}{\surd(2\pi)}\cdot\left\lvert \dfrac{1}{2\surd y}\right\rvert~\mathbf 1_{y\in[0..\infty)}+\dfrac{\mathrm e^{-y/2}}{\surd(2\pi)}\cdot\left\lvert \dfrac{-1}{2\surd y}\right\rvert~\mathbf 1_{y\in(0..\infty)}
\\[1ex]&= \dfrac{\mathrm e^{-y/2}}{\surd(2\pi y)}~\mathbf 1_{y\in(0..\infty)} \underbrace{+ \dfrac{\mathrm e^{-y/2}}{2\surd(2\pi y)}~\mathbf 1_{y=0}}_{\text{which we usually ignore}\\{\tiny\text{because its just one point}}}
\end{align}$$
