Answer
$-\dfrac{x}{x+2}; x \ne -2, 1$
Work Step by Step
The given expression is equivalent to:
$=\dfrac{-x^2+x}{x^2+x-2}$
Factor the numerator and the denominator completely to obtain:
$=\dfrac{-x(x-1)}{(x+2)(x-1)}$
Cancel the common factors to obtain:
$\require{cancel}
\\=\dfrac{-x\cancel{(x-1)}}{(x+2)\cancel{(x-1)}}
\\=-\dfrac{x}{x+2}; x \ne -2, 1$
($x$ cannot be $-2$ and $1$ because they will make the original expression undefined.)
