Answer
$$\frac{1}{x}$$
Work Step by Step
$$\eqalign{
& y = \ln 3x \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\ln 3x} \right] \cr
& {\text{use the formula }}\cr
& \frac{d}{{dx}}\ln u = \frac{1}{u}\frac{{du}}{{dx}}{\text{, }}u \gt 0\cr
& {\text{ where }}u{\text{ is any differentiable function of }}x \cr
& {\text{For this exercise you can note that }}u = 3x;{\text{ then}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{3x}}\frac{d}{{dx}}\left[ {3x} \right] \cr
& {\text{solve the derivative and simplify}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{3x}}\left( 3 \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{x} \cr} $$
