Let $c >0 $ and $ f:(0,\infty) \to \mathbb R, f(x)=\frac{2x}{c}e^{-x^2/c}$ Let $X_c$ be a related random variable. I have shown that $f$ is a density. I want to calculate the distribution $W=e^{-X_c^2/c}$ $$P(e^{-X_c^2/c} \le w)=P(ln(\frac{1}{w})c \le X_c^2)=1-P(X_c \le \sqrt{ln( \frac{1}{w})c}$$ So $$f_w-1=f_c(\sqrt{ln( \frac{1}{w})c}) \cdot \frac{-c}{2w\sqrt{-c \cdot ln( w)}}=...=-e^{-ln(\frac{1}{w})\cdot c}=\frac{-w}{c}$$
And so i get $f_w=1-\frac{-w}{c}$
P.S. the derivative from $\sqrt{ln(\frac{1}{w})c} $ is $\frac{-c}{2w \sqrt{-c \cdot ln( w)}}$. The problem I have is the negative root.How do I fix this? Any help is much appreciated!
Note that $e^{-X^2_c/c}$ is always in $[0,1]$. So you should only consider $P(e^{-X^2_c/c} \le w)$ for $w \in [0,1]$ anyway.
I think the CDF of $X_c$ can be shown to be $F_c(x) = 1-e^{-x^2/c}$ for $x \ge 0$. Thus your first line shows $$P(e^{-X^2_c / c} \le w) = 1 - P(X_c \le \sqrt{c \ln(1/w)}) = w, \qquad 0 \le w \le 1.$$
Thus $e^{-X^2_c/c} \sim \text{Uniform}([0,1]).$