Tomamos como el límite
$$\lim_{x \to 3^+}\left(\frac{- 3 x + \left(x^{3} + 2\right)}{- 4 x + \left(x^{2} + 3\right)}\right)$$
cambiamos
$$\lim_{x \to 3^+}\left(\frac{- 3 x + \left(x^{3} + 2\right)}{- 4 x + \left(x^{2} + 3\right)}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{\left(x - 1\right)^{2} \left(x + 2\right)}{\left(x - 3\right) \left(x - 1\right)}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{\left(x - 1\right) \left(x + 2\right)}{x - 3}\right) = $$
False
= oo
Entonces la respuesta definitiva es:
$$\lim_{x \to 3^+}\left(\frac{- 3 x + \left(x^{3} + 2\right)}{- 4 x + \left(x^{2} + 3\right)}\right) = \infty$$