Tomamos como el límite
$$\lim_{x \to -1^+}\left(\frac{- 3 x + \left(x^{3} - 2\right)}{- x^{2} + x}\right)$$
cambiamos
$$\lim_{x \to -1^+}\left(\frac{- 3 x + \left(x^{3} - 2\right)}{- x^{2} + x}\right)$$
=
$$\lim_{x \to -1^+}\left(\frac{\left(x - 2\right) \left(x + 1\right)^{2}}{\left(-1\right) x \left(x - 1\right)}\right)$$
=
$$\lim_{x \to -1^+}\left(\frac{\left(2 - x\right) \left(x + 1\right)^{2}}{x \left(x - 1\right)}\right) = $$
$$\frac{\left(-1 + 1\right)^{2} \left(2 - -1\right)}{\left(-1\right) \left(-1 - 1\right)} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to -1^+}\left(\frac{- 3 x + \left(x^{3} - 2\right)}{- x^{2} + x}\right) = 0$$