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