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