TeX source:
\begin{array}{lclcrcl}\quad\mathbb{D}&=&\mathbb{R}\setminus_{\{1\}} \cr \cr f(x) &=& \dfrac{x^2+1}{x^3-1} \cr \cr \quad u(x) &=& x^2+1 & \Rightarrow & u'(x) &=& 2x \cr \quad v(x) &=& x^3-1 & \Rightarrow & v'(x) &=& 3x^2 \cr \cr f'(x) &=& \dfrac{2x \cdot (x^3-1)-(x^2+1) \cdot 3x^2}{(x^3-1)^2} \cr\cr &=& \dfrac{2x^4-2x-3x^4-3x^2}{x^6-2x^3+1} \cr\cr &=& \dfrac{-x^4-3x^2-2x}{x^6-2x^3+1} \end{array}