Answer
$$\frac{{dy}}{{dx}} = \frac{{{{\left( {3x + 2} \right)}^{2/3}}}}{{3x - 1}}\left( {\frac{2}{{3x + 2}} - \frac{1}{{3x - 1}}} \right)$$
Work Step by Step
$$\eqalign{
& y = \frac{{{{\left( {3x + 2} \right)}^{2/3}}}}{{3x - 1}} \cr
& {\text{Differentiate by logarithmic differentiation}} \cr
& {\text{Take the natural logarithm of both sides}} \cr
& \ln y = \ln \left( {\frac{{{{\left( {3x + 2} \right)}^{2/3}}}}{{3x - 1}}} \right) \cr
& {\text{Use the property }}\ln \left( {\frac{a}{b}} \right) = \ln a - \ln b \cr
& \ln y = \ln {\left( {3x + 2} \right)^{2/3}} - \ln \left( {3x - 1} \right) \cr
& {\text{Use the property }}\ln \left( {{a^n}} \right) = n\ln a \cr
& \ln y = \frac{2}{3}\ln \left( {3x + 2} \right) - \ln \left( {3x - 1} \right) \cr
& {\text{Differentiate both sides w}}{\text{.r}}{\text{.t}} \cr
& \frac{d}{{dx}}\left[ {\ln y} \right] = \frac{2}{3}\frac{d}{{dx}}\left[ {\ln \left( {3x + 2} \right)} \right] - \frac{d}{{dx}}\left[ {\ln \left( {3x - 1} \right)} \right] \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{2}{3}\left( {\frac{3}{{3x + 2}}} \right) - \frac{1}{{3x - 1}} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{2}{{3x + 2}} - \frac{1}{{3x - 1}} \cr
& \frac{{dy}}{{dx}} = y\left( {\frac{2}{{3x + 2}} - \frac{1}{{3x - 1}}} \right) \cr
& {\text{Substitute back }}y = \frac{{{{\left( {3x + 2} \right)}^{2/3}}}}{{3x - 1}} \cr
& \frac{{dy}}{{dx}} = \frac{{{{\left( {3x + 2} \right)}^{2/3}}}}{{3x - 1}}\left( {\frac{2}{{3x + 2}} - \frac{1}{{3x - 1}}} \right) \cr} $$
