Answer
$$\frac{{dy}}{{dx}} = \frac{{{x^2}{{\left( {3x + 1} \right)}^2}}}{{{{\left( {2x - 1} \right)}^3}}}\left( {\frac{2}{x} + \frac{6}{{3x + 1}} - \frac{6}{{2x - 1}}} \right)$$
Work Step by Step
$$\eqalign{
& y = \frac{{{x^2}{{\left( {3x + 1} \right)}^2}}}{{{{\left( {2x - 1} \right)}^3}}} \cr
& {\text{Differentiate by logarithmic differentiation}} \cr
& {\text{Take the natural logarithm of both sides}} \cr
& \ln y = \ln \left( {\frac{{{x^2}{{\left( {3x + 1} \right)}^2}}}{{{{\left( {2x - 1} \right)}^3}}}} \right) \cr
& {\text{Use the property }}\ln \left( {\frac{a}{b}} \right) = \ln a - \ln b \cr
& \ln y = \ln {x^2}{\left( {3x + 1} \right)^2} - \ln {\left( {2x - 1} \right)^3} \cr
& {\text{Use the property }}\ln \left( {ab} \right) = \ln a + \ln b \cr
& \ln y = \ln {x^2} + \ln {\left( {3x + 1} \right)^2} - \ln {\left( {2x - 1} \right)^3} \cr
& {\text{Use the powe property }}\ln \left( {{a^n}} \right) = n\ln a \cr
& \ln y = 2\ln x + 2\ln \left( {3x + 1} \right) - 3\ln \left( {2x - 1} \right) \cr
& {\text{Differentiate both sides w}}{\text{.r}}{\text{.t}} \hspace{2mm} x \cr
& \frac{d}{{dx}}\left[ {\ln y} \right] = \frac{d}{{dx}}\left[ {2\ln x} \right] + \frac{d}{{dx}}\left[ {2\ln \left( {3x + 1} \right)} \right] - \frac{d}{{dx}}\left[ {3\ln \left( {2x - 1} \right)} \right] \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{2}{x} + \frac{6}{{3x + 1}} - \frac{6}{{2x - 1}} \cr
& {\text{Solve for }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = y\left( {\frac{2}{x} + \frac{6}{{3x + 1}} - \frac{6}{{2x - 1}}} \right) \cr
& {\text{Substitute back }}y = \frac{{{x^2}{{\left( {3x + 1} \right)}^2}}}{{{{\left( {2x - 1} \right)}^3}}} \cr
& \frac{{dy}}{{dx}} = \frac{{{x^2}{{\left( {3x + 1} \right)}^2}}}{{{{\left( {2x - 1} \right)}^3}}}\left( {\frac{2}{x} + \frac{6}{{3x + 1}} - \frac{6}{{2x - 1}}} \right) \cr} $$
