I'm sure this is a trivial problem, but I am having trouble committing to an answer.
If $U$~$(0,1)$, then $Y=-ln U$. I know that $\mathbb P(Y \le y)$
= $\mathbb P(g(X) \le y)$
=$\mathbb P(-lnU \le y)$
=$\mathbb P(U \ge e^{-y})$
So would my CDF be $F_Y (y)=e^{-y},0\le y \le \infty$? Then of course, my PDF would be $-e^{-y}$.
It is an exponential distribution, but your derivation contains a mistake.
$\mathbb P(U \ge e^{-y}) = e^{-y}$, as $U$ is uniform on $(0,1)$. Then, $F_Y(y) = 1-e^{-y}$ and the PDF is $f_Y(y) = e^{-y}$.
You can use your intuition with simple expressions. The expression you obtained for $F$ is decreasing and the expression for the PDF is negative, and we know that CDFs are increasing, and that PDFs are positive.
You can find this in Wikipedia:
https://en.wikipedia.org/wiki/Exponential_distribution#Generating_exponential_variates
