Answer
$$ y^{\prime}=\frac{1}{3} \sqrt[3]{\frac{x(x-2)}{x^{2}+1}}\left(\frac{1}{x}+\frac{1}{x-2}-\frac{2 x}{x^{2}+1}\right) $$
Work Step by Step
Given $$ y=\sqrt[3]{\frac{x(x-2)}{x^{2}+1}}$$
So, we have
\begin{aligned}
&\Rightarrow \ln y=\ln\sqrt[3]{\frac{x(x-2)}{x^{2}+1}} \\
&\Rightarrow \ln y= \frac{1}{3}\ln \frac{x(x-2)}{x^{2}+1} \\
&\Rightarrow \ln y=\frac{1}{3}\left[\ln x (x-2)-\ln \left(x^{2}+1\right)\right]\\
&\Rightarrow \ln y=\frac{1}{3}\left[\ln x+\ln (x-2)-\ln \left(x^{2}+1\right)\right]\\
&\text{differentiate both sides with respect to } x\\
& \Rightarrow \frac{y^{\prime}}{y}=\frac{1}{3}\left(\frac{1}{x}+\frac{1}{x-2}-\frac{2 x}{x^{2}+1}\right) \\
&
\Rightarrow y^{\prime}=\frac{y}{3}\left(\frac{1}{x}+\frac{1}{x-2}-\frac{2 x}{x^{2}+1}\right) \\
&
\Rightarrow y^{\prime}=\frac{1}{3} \sqrt[3]{\frac{x(x-2)}{x^{2}+1}}\left(\frac{1}{x}+\frac{1}{x-2}-\frac{2 x}{x^{2}+1}\right)
\end{aligned}
