Answer
$$y'=\frac{1}{1+\theta\ln3}$$
Work Step by Step
$$y=\log_3(1+\theta\ln3)$$
Recall Theorem 7: $$\frac{d}{d\theta}\log_au=\frac{1}{u\ln a}\frac{du}{d\theta}$$
Apply here to find $y'$: $$y'=(\log_3(1+\theta\ln3))'=\frac{1}{(1+\theta\ln3)\ln3}(1+\theta\ln3)'$$
$$y'=\frac{(0+1\times\ln3)}{(1+\theta\ln3)\ln3}=\frac{\ln3}{(1+\theta\ln3)\ln3}$$
$$y'=\frac{1}{1+\theta\ln3}$$