The region of integration looks like as follows:
I started by computing the jacobian which I found to be $|J(\rho,\theta)|=\frac{-e^{-2\rho}}{6}$. And with the new variables the integrand becomes $\frac{-e^{-2\rho}}{6}ln(e^{-2\rho})=\rho \frac{e^{-2\rho}}{3}$
However, I can't figure out how to determine the bounds for $\rho$ and $\theta$. If we were to compute this integral in cartesian coordinates we would need to compute two double integrals and subtract one from the other. I saw a different question on math MSE involving an general elliptical region with a and b axes and the substitution $x=arcos(\theta)$ and $y=brsin(\theta)$ $\theta \in[0,2\pi]$ and $r\in[0,1]$ was suggested in the answer.
So for this question should we define $e^{-\rho}\in[0,1]$ implying $\rho\in[0,\infty)$? Is this correct? But what would make use choose $e^{-\rho}\in[0,1]$ in the first place?
Your neww coordinates are $(\rho,\theta)$, with are polar coordinates for the annulus. Think of it like you are integrating over all the circles contained in the annulus, that's it, circunferences with $\rho\in(1,2)$ with $\theta\in (0,2\pi)$.