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