Answer
$0$
Work Step by Step
Factor each polynomial:
$$\lim_{x\to -1^-}\frac{x^2-1}{x^3-1}\\=\lim_{x\to -1^-}\frac{(x+1)(x-1)}{(x-1)(x^2+x+1)}.$$
Cancel the common factors: $$\lim_{x\to -1^-}\frac{x+1}{x^2+x+1}\\=\frac{(-1)+1}{(-1)^2+(-1)+1}\\=\frac{0}{1}=0$$