Answer
$$\frac{{1 - 3{v^2}}}{{3{v^{2/3}}{{\left( {3{v^2} + 2v + 1} \right)}^{4/3}}}}$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dv}}{\left( {\frac{v}{{3{v^2} + 2v + 1}}} \right)^{1/3}} \cr
& {\text{using the chain rule}} \cr
& = \frac{1}{3}{\left( {\frac{v}{{3{v^2} + 2v + 1}}} \right)^{1/3 - 1}}\frac{d}{{dv}}\left( {\frac{v}{{3{v^2} + 2v + 1}}} \right) \cr
& {\text{use the quotient rule}} \cr
& = \frac{1}{3}{\left( {\frac{v}{{3{v^2} + 2v + 1}}} \right)^{ - 2/3}}\left( {\frac{{\left( {3{v^2} + 2v + 1} \right)\frac{d}{{dv}}\left[ v \right] - v\frac{d}{{dv}}\left[ {3{v^2} + 2v + 1} \right]}}{{{{\left( {3{v^2} + 2v + 1} \right)}^2}}}} \right) \cr
& {\text{solving derivatives}} \cr
& = \frac{1}{3}{\left( {\frac{v}{{3{v^2} + 2v + 1}}} \right)^{ - 2/3}}\left( {\frac{{\left( {3{v^2} + 2v + 1} \right)\left( 1 \right) - v\left( {6v + 2} \right)}}{{{{\left( {3{v^2} + 2v + 1} \right)}^2}}}} \right) \cr
& {\text{multiplying}} \cr
& = \frac{1}{3}{\left( {\frac{v}{{3{v^2} + 2v + 1}}} \right)^{ - 2/3}}\left( {\frac{{3{v^2} + 2v + 1 - 6{v^2} - 2v}}{{{{\left( {3{v^2} + 2v + 1} \right)}^2}}}} \right) \cr
& = \frac{1}{3}\left( {\frac{{{v^{ - 2/3}}}}{{{{\left( {3{v^2} + 2v + 1} \right)}^{ - 2/3}}}}} \right)\left( {\frac{{1 - 3{v^2}}}{{{{\left( {3{v^2} + 2v + 1} \right)}^2}}}} \right) \cr
& = \frac{1}{3}\left( {\frac{{{v^{ - 2/3}}\left( {1 - 3{v^2}} \right)}}{{{{\left( {3{v^2} + 2v + 1} \right)}^{ - 2/3 + 2}}}}} \right) \cr
& = \frac{1}{3}\left( {\frac{{{v^{ - 2/3}}\left( {1 - 3{v^2}} \right)}}{{{{\left( {3{v^2} + 2v + 1} \right)}^{4/3}}}}} \right) \cr
& = \frac{{1 - 3{v^2}}}{{3{v^{2/3}}{{\left( {3{v^2} + 2v + 1} \right)}^{4/3}}}} \cr} $$
