## Calculus (3rd Edition)

$$y'=\frac{1}{x+1}- \frac{3x^2}{x^3+1}.$$
Since $y=\ln\left( \frac{x+1}{x^3+1}\right)$, first we simplify $y$ as follows $$y=\ln (x+1)- \ln (x^3+1)$$ then we have $$y'=\frac{1}{x+1}- \frac{3x^2}{x^3+1}.$$