Answer
\[\frac{{dy}}{{dx}} = \frac{{2x - 1}}{{3\,{{\left( {{x^2} - x + 1} \right)}^{\frac{2}{3}}}}}\]
Work Step by Step
\[\begin{gathered}
y = \sqrt[3]{{{x^2} - x + 1}} \hfill \\
\hfill \\
rewrite\, \hfill \\
\hfill \\
y = \,{\left( {{x^2} - x + 1} \right)^{\frac{1}{3}}} \hfill \\
\hfill \\
differentiate \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{1}{3}\,{\left( {{x^2} - x + 1} \right)^{ - \frac{2}{3}}}\,\left( {2x - 1} \right) \hfill \\
\hfill \\
use\,\,{x^{ - m}} = \frac{1}{{{x^m}}} \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{{2x - 1}}{{3\,{{\left( {{x^2} - x + 1} \right)}^{\frac{2}{3}}}}} \hfill \\
\end{gathered} \]
