Tomamos como el límite
$$\lim_{x \to 3^+}\left(\frac{x^{2} + \left(x - 6\right)}{- x + \left(x^{3} - 12\right)}\right)$$
cambiamos
$$\lim_{x \to 3^+}\left(\frac{x^{2} + \left(x - 6\right)}{- x + \left(x^{3} - 12\right)}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{\left(x - 2\right) \left(x + 3\right)}{x^{3} - x - 12}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{x^{2} + x - 6}{x^{3} - x - 12}\right) = $$
$$\frac{-6 + 3 + 3^{2}}{-12 - 3 + 3^{3}} = $$
= 1/2
Entonces la respuesta definitiva es:
$$\lim_{x \to 3^+}\left(\frac{x^{2} + \left(x - 6\right)}{- x + \left(x^{3} - 12\right)}\right) = \frac{1}{2}$$