Answer
$$\frac{{dy}}{{dx}} = \frac{{2x + 6 - \left( {2x - 1} \right)\ln \left| {2x - 1} \right|}}{{\left( {2x - 1} \right){{\left( {x + 3} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{\ln \left| {2x - 1} \right|}}{{x + 3}} \cr
& {\text{differentiate both sides}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {\frac{{\ln \left| {2x - 1} \right|}}{{x + 3}}} \right) \cr
& {\text{use quotient rule}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {x + 3} \right)\frac{d}{{dx}}\left( {\ln \left| {2x - 1} \right|} \right) - \left( {\ln \left| {2x - 1} \right|} \right)\frac{d}{{dx}}\left( {x + 3} \right)}}{{{{\left( {x + 3} \right)}^2}}} \cr
& {\text{use }}\frac{d}{{dx}}\ln g\left( x \right) = \frac{1}{{g\left( x \right)}}g'\left( x \right) \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {x + 3} \right)\left( {\frac{2}{{2x - 1}}} \right) - \left( {\ln \left| {2x - 1} \right|} \right)\left( 1 \right)}}{{{{\left( {x + 3} \right)}^2}}} \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{{2x + 6}}{{2x - 1}} - \ln \left| {2x - 1} \right|}}{{{{\left( {x + 3} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{2x + 6 - \left( {2x - 1} \right)\ln \left| {2x - 1} \right|}}{{\left( {2x - 1} \right){{\left( {x + 3} \right)}^2}}} \cr} $$
