Tomamos como el límite
$$\lim_{x \to -4^+}\left(\frac{64 - x^{3}}{3 x + \left(2 - 2 x^{2}\right)}\right)$$
cambiamos
$$\lim_{x \to -4^+}\left(\frac{64 - x^{3}}{3 x + \left(2 - 2 x^{2}\right)}\right)$$
=
$$\lim_{x \to -4^+}\left(\frac{\left(-1\right) \left(x - 4\right) \left(x^{2} + 4 x + 16\right)}{\left(-1\right) \left(x - 2\right) \left(2 x + 1\right)}\right)$$
=
$$\lim_{x \to -4^+}\left(\frac{x^{3} - 64}{2 x^{2} - 3 x - 2}\right) = $$
$$\frac{-64 + \left(-4\right)^{3}}{-2 - -12 + 2 \left(-4\right)^{2}} = $$
= -64/21
Entonces la respuesta definitiva es:
$$\lim_{x \to -4^+}\left(\frac{64 - x^{3}}{3 x + \left(2 - 2 x^{2}\right)}\right) = - \frac{64}{21}$$