Tomamos como el límite
$$\lim_{x \to -3^+}\left(\frac{\left(9 - x^{2}\right) \left(x + 3\right)}{2 - x}\right)$$
cambiamos
$$\lim_{x \to -3^+}\left(\frac{\left(9 - x^{2}\right) \left(x + 3\right)}{2 - x}\right)$$
=
$$\lim_{x \to -3^+}\left(\frac{\left(-1\right) \left(x - 3\right) \left(x + 3\right)^{2}}{2 - x}\right)$$
=
$$\lim_{x \to -3^+}\left(\frac{\left(x - 3\right) \left(x + 3\right)^{2}}{x - 2}\right) = $$
$$\frac{\left(-3 - 3\right) \left(-3 + 3\right)^{2}}{-3 - 2} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to -3^+}\left(\frac{\left(9 - x^{2}\right) \left(x + 3\right)}{2 - x}\right) = 0$$