Tomamos como el límite
$$\lim_{x \to -3^+}\left(\frac{- 6 x^{2} + \left(x^{4} - 27\right)}{- 3 x^{2} + \left(x^{3} + \left(x + 3\right)\right)}\right)$$
cambiamos
$$\lim_{x \to -3^+}\left(\frac{- 6 x^{2} + \left(x^{4} - 27\right)}{- 3 x^{2} + \left(x^{3} + \left(x + 3\right)\right)}\right)$$
=
$$\lim_{x \to -3^+}\left(\frac{\left(x - 3\right) \left(x + 3\right) \left(x^{2} + 3\right)}{x^{3} - 3 x^{2} + x + 3}\right)$$
=
$$\lim_{x \to -3^+}\left(\frac{x^{4} - 6 x^{2} - 27}{x^{3} - 3 x^{2} + x + 3}\right) = $$
$$\frac{- 6 \left(-3\right)^{2} - 27 + \left(-3\right)^{4}}{\left(-3\right)^{3} - 3 \left(-3\right)^{2} - 3 + 3} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to -3^+}\left(\frac{- 6 x^{2} + \left(x^{4} - 27\right)}{- 3 x^{2} + \left(x^{3} + \left(x + 3\right)\right)}\right) = 0$$