Answer
$$f'\left( x \right) = \frac{{12{{\left( {x - 1} \right)}^2}}}{{{{\left( {3x + 1} \right)}^4}}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\left[ {\frac{{x - 1}}{{3x + 1}}} \right]^3} \cr
& {\text{Differentiate both sides with respect to }}x \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{{\left( {\frac{{x - 1}}{{3x + 1}}} \right)}^3}} \right] \cr
& {\text{Apply the chain rule and the powe rule }}\frac{d}{{dx}}\left[ {{u^n}} \right] = n{u^{n - 1}}u' \cr
& {\text{For this exercise, }}u = \frac{{x - 1}}{{3x + 1}}{\text{ and }}n = 3,{\text{ then}} \cr
& f'\left( x \right) = 3{\left( {\frac{{x - 1}}{{3x + 1}}} \right)^{3 - 1}}\underbrace {\frac{d}{{dx}}\left[ {\frac{{x - 1}}{{3x + 1}}} \right]}_{{\text{Use quotient rule}}} \cr
& f'\left( x \right) = 3{\left( {\frac{{x - 1}}{{3x + 1}}} \right)^2}\left[ {\frac{{\left( {3x + 1} \right)\frac{d}{{dx}}\left[ {x - 1} \right] - \left( {x - 1} \right)\frac{d}{{dx}}\left[ {3x + 1} \right]}}{{{{\left( {3x + 1} \right)}^2}}}} \right] \cr
& {\text{Computing derivatives}} \cr
& f'\left( x \right) = 3{\left( {\frac{{x - 1}}{{3x + 1}}} \right)^2}\left[ {\frac{{\left( {3x + 1} \right)\left( 1 \right) - \left( {x - 1} \right)\left( 3 \right)}}{{{{\left( {3x + 1} \right)}^2}}}} \right] \cr
& {\text{Simplifying}} \cr
& f'\left( x \right) = 3{\left( {\frac{{x - 1}}{{3x + 1}}} \right)^2}\left[ {\frac{{3x + 1 - 3x + 3}}{{{{\left( {3x + 1} \right)}^2}}}} \right] \cr
& f'\left( x \right) = 3{\left( {\frac{{x - 1}}{{3x + 1}}} \right)^2}\left[ {\frac{4}{{{{\left( {3x + 1} \right)}^2}}}} \right] \cr
& f'\left( x \right) = \frac{{12{{\left( {x - 1} \right)}^2}}}{{{{\left( {3x + 1} \right)}^4}}} \cr} $$
