Answer
$$\frac{{dy}}{{dx}} = \frac{{4x}}{3}\root 3 \of {\frac{{{x^2} - 1}}{{{x^2} + 1}}} \left( {\frac{1}{{{x^4} - 1}}} \right)$$
Work Step by Step
$$\eqalign{
& y = \root 3 \of {\frac{{{x^2} - 1}}{{{x^2} + 1}}} \cr
& {\text{Rewrite the radical}} \cr
& y = {\left( {\frac{{{x^2} - 1}}{{{x^2} + 1}}} \right)^{1/3}} \cr
& {\text{Take natural logarithm on both sides}} \cr
& \ln y = \ln {\left( {\frac{{{x^2} - 1}}{{{x^2} + 1}}} \right)^{1/3}} \cr
& {\text{Using the logarithmic properties}} \cr
& \ln y = \frac{1}{3}\ln \left( {\frac{{{x^2} - 1}}{{{x^2} + 1}}} \right) \cr
& \ln y = \frac{1}{3}\ln \left( {{x^2} - 1} \right) - \frac{1}{3}\ln \left( {{x^2} + 1} \right) \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{3}\left( {\frac{{2x}}{{{x^2} - 1}}} \right) - \frac{1}{3}\left( {\frac{{2x}}{{{x^2} + 1}}} \right) \cr
& {\text{Solving for }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{{2y}}{3}\left[ {\frac{x}{{{x^2} - 1}} - \frac{x}{{{x^2} + 1}}} \right] \cr
& \frac{{dy}}{{dx}} = \frac{{2y}}{3}\left( {\frac{{{x^3} + x - {x^3} + x}}{{{x^4} - 1}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{{2y}}{3}\left( {\frac{{2x}}{{{x^4} - 1}}} \right) \cr
& {\text{Substituting }}y = \root 3 \of {\frac{{{x^2} - 1}}{{{x^2} + 1}}} \cr
& \frac{{dy}}{{dx}} = \frac{{4x}}{3}\root 3 \of {\frac{{{x^2} - 1}}{{{x^2} + 1}}} \left( {\frac{1}{{{x^4} - 1}}} \right) \cr} $$
