Answer
$$\frac{{dy}}{{d\theta }} = \frac{1}{{1 + \theta \ln 3}}$$
Work Step by Step
$$\eqalign{
& y = {\log _3}\left( {1 + \theta \ln 3} \right) \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}\theta \cr
& \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {{{\log }_3}\left( {1 + \theta \ln 3} \right)} \right] \cr
& {\text{use the rule }}\frac{d}{{dx}}\left[ {{{\log }_a}u} \right] = \frac{1}{{\ln a}} \cdot \frac{1}{u}\frac{{du}}{{dx}}. \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{\ln 3}} \cdot \frac{1}{{1 + \theta \ln 3}}\frac{d}{{d\theta }}\left[ {1 + \theta \ln 3} \right] \cr
& {\text{solve the derivative}} \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{\ln 3}} \cdot \frac{1}{{1 + \theta \ln 3}}\left( {\ln 3} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{1 + \theta \ln 3}} \cr} $$
