Calculus (3rd Edition)

$$y'(x)= 1-2x$$
Given $$x+y x^{-1}=1\ \tag{1}$$$$y=x-x^{2} \tag{2}$$ From (1), multiply by $x$ \begin{align*} x+y x^{-1}&=1\\ x^2+y&=x\\ y&= x-x^2 \end{align*} which is the same as the second equation. Differentiating (1), we get \begin{align*} 1-y x^{-2}+x^{-1} y^{\prime}&=0\\ y'(x) &= \frac{y x^{-2}-1}{x^{-1}}\\ &= y x^{-1}-x \end{align*} Put (2) in the last equation to get $$y'(x)= 1-2x$$ which is the derivative of the second equation.