Sorry expression for the PDF(x) did not appear it is PDF for Gaussian distribution.
I think that
if PDF(x) is Gaussian and f(x)=x^2;
then PDF_x^2(x)=const*PDF(sqrt(x)) so the x should be replaced by sqrt(x) in Gaussian PDF in case of 0 mean. In case non 0 man x-mean should be replaced by sqrt(x-mean).
You are partially right. ). When can we say that and therefore ? This holds, in particular, when f increases monotonically. Then f is injective, so exists.
If X is non-negative, then because increases monotonically when . However, when X is normal, it can assume negative values. In this case, . If , this equals . In general, if , you can use the fact that where is the PDF of .
Sorry, I have not been studying or teaching probability lately, so I don't know good textbooks. But this is very basic stuff: just the definition of PDF and properties of normal distribution. I think that every textbook on probability should cover this.