Answer
$$\frac{{dy}}{{dx}} = \frac{{x - 3 - x\ln \left| {3x} \right|}}{{x{{\left( {x - 3} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{\ln \left| {3x} \right|}}{{x - 3}} \cr
& {\text{differentiate both sides}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{\ln \left| {3x} \right|}}{{x - 3}}} \right] \cr
& {\text{by using the quotient rule}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {x - 3} \right) \cdot {D_x}\left( {\ln \left| {3x} \right|} \right) - \ln 3x \cdot {D_x}\left( {x - 3} \right)}}{{{{\left( {x - 3} \right)}^2}}} \cr
& {\text{use }}\frac{d}{{dx}}\ln g\left( x \right) = \frac{1}{{g\left( x \right)}}g'\left( x \right) \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {x - 3} \right)\left( {1/x} \right) - \ln \left| {3x} \right|\left( 1 \right)}}{{{{\left( {x - 3} \right)}^2}}} \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{{x - 3 - x\ln \left| {3x} \right|}}{x}}}{{{{\left( {x - 3} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{x - 3 - x\ln \left| {3x} \right|}}{{x{{\left( {x - 3} \right)}^2}}} \cr} $$
