I'm trying to find the PDF of the product of two random variables by first finding the CDF. I don't know where I'm going wrong. Let $X\sim N(0,1)$ and $Y\sim Uniform\{-1,1\}$ and let $Z = XY$, then:
$F_Z(Z<z) = P(Z<z) = P(XY<z) = P(Y<\frac{z}{X})$
$\Rightarrow F_Z(Z<z) = > \displaystyle\int_{-\infty}^{\infty}{F_Y\left(\frac{z}{x}\right)f_X(x)dx}$
$\Rightarrow F_Z(Z<z) = > \displaystyle\int_{-\infty}^{-z}{F_Y\left(\frac{z}{x}\right)f_X(x)dx}+\displaystyle\int_{-z}^{0}{F_Y\left(\frac{z}{x}\right)f_X(x)dx}+\displaystyle\int_{-0}^{z}{F_Y\left(\frac{z}{x}\right)f_X(x)dx}+\displaystyle\int_{z}^{\infty}{F_Y\left(\frac{z}{x}\right)f_X(x)dx}$
$\Rightarrow F_Z(Z<z) = > \displaystyle\int_{-\infty}^{-z}{\left(\frac{\frac{z}{x}+1}{2}\right)f_X(x)dx}+\displaystyle\int_{0}^{z}{f_X(x)dx}+\displaystyle\int_{z}^{\infty}{\left(\frac{\frac{z}{x}+1}{2}\right)f_X(x)dx}$
$\Rightarrow F_Z(Z<z) = > \displaystyle\int_{-\infty}^{-z}{\left(\frac{z}{2x}\right)f_X(x)dx}+\displaystyle\int_{z}^{\infty}{\left(\frac{z}{2x}\right)f_X(x)dx}+\displaystyle\int_{0}^{z}{f_X(x)dx}+\displaystyle\int_{z}^{\infty}{\left(\frac{1}{2}\right)f_X(x)dx}+\displaystyle\int_{-\infty}^{-z}{\left(\frac{1}{2}\right)f_X(x)dx}$
Since $f_X(x)$ is an even function.
$F_Z(Z<z) = \displaystyle\int_{0}^{\infty}{f_X(x)dx}$
I want to know where am I going wrong.
As pointed out in other answer, you need to be careful with passing terms in inequalities when sign can be negative. I would write:
$$P(Z \le z)= \int f_X(x) P(XY \le z \mid X=x)\,dx=\\ = \int_{x<0} f_X(x) P(Y \ge \frac{z}{x}) \,dx + \int_{x\ge 0} f_X(x) P(Y\le \frac{z}{x})\,dx $$
Can you go on from here?