probability - find PDF of a random variable

Let $X$ be a continuous random variable with PDF $$ f_X(x) = \frac 1 {\sqrt{2\pi}} e^{-\frac{x^2}{2}} \qquad \forall x \in \mathbb{R}. $$ and let $ Y = \sqrt{|X|}$, find $f_Y(y)$.

I know that that $x_1 = y^2$ and $x_2 = -y^2$

And $g’(x) = \frac{x}{2|x|^{3/2}}$

I can't find the PDF. I've tried these formulas:
$$f_Y(y) = f_X(g(x_1)) \cdot \left| g'(x_1) \right| + f_X(g(x_2)) \cdot \left| g'(x_2) \right|$$

$$f_Y(y) = \sum_{I=1}^n \frac{f_X(x)}{|g'(x)|}$$
$$ f_Y(y) = f_X(y^2)+ f_X(-y^2) $$

I can't figure out what formula to use, because every site i'm checking is saying another formula, I know the result is: $4y\frac{1}{\sqrt{2\pi}}e^{-\frac{y^4}{2}}$

Answer

To find the pdf of a continuous random variable, it is more general to start with finding its cdf, from which then taking the derivative. It's more useful for transformations such as in your question, for which a direct Jacobian is not available due to the non-differentiability of the absolute value function.

Clearly, $Y$ has range $[0, +\infty)$, so for each $y \in (0, +\infty)$,
we have
\begin{align*}

F_Y(y) =& P[Y \leq y] = P[\sqrt{|X|} \leq y] = P[|X| \leq y^2] \\
=& P[-y^2 \leq X \leq y^2] \\
=& \Phi(y^2) - \Phi(-y^2) \\
=& 2\Phi(y^2) - 1,
\end{align*}
where $\Phi(\cdot)$ is the cdf of standard normal distribution. By convention, we use $\phi$ (i.e., $f_X$ in your question) to denote the density of standard normal. It follows that the density of $Y$ can be obtained by
\begin{align}
f_Y(y) = F_Y'(y) = 2\phi(y^2)\times 2y = 4y\phi(y^2) = \frac{4y}{\sqrt{2\pi}}e^{-\frac{1}{2}y^4}, \quad 0 < y < \infty.
\end{align}