Answer
$$ y'(x)= 1-2x $$
Work Step by Step
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.
