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