Answer
$\dfrac{-x}{x+2}, \quad x\ne-2, 1$
Work Step by Step
By factoring both denominator and numerator and canceling the common factors, we obtain:
\begin{align*}
\require{cancel}
\dfrac{x-x^2}{x^2+x-2}&=\dfrac{-x(x-1)}{(x+2)(x-1)}\\
\\&=\dfrac{-x\cancel{(x-1)}}{(x+2)\cancel{(x-1)}}\\
\\&=\dfrac{-x}{x+2}, \quad x\ne-2, 1
\end{align*}
