Answer
$$y'=\frac{1-\theta}{\theta}$$
Work Step by Step
$$y=\ln(3\theta e^{-\theta})$$
We have $$(\ln \theta)'=\frac{1}{\theta}$$ $$(e^{\theta})'=e^\theta$$
So the derivative of $y$ is $$y'=\frac{1}{3\theta e^{-\theta}}(3\theta e^{-\theta})'=\frac{3(e^{-\theta}+\theta e^{-\theta}(-\theta)')}{3\theta e^{-\theta}}$$
$$y'=\frac{e^{-\theta}-\theta e^{-\theta}}{\theta e^{-\theta}}=\frac{e^{-\theta}(1-\theta)}{\theta e^{-\theta}}$$
$$y'=\frac{1-\theta}{\theta}$$
