We are given the limit:

$\lim_{{x \to -1}} \left( \frac{1}{x + 1} - \frac{3}{x^3 + 1} \right)$

Step-by-step solution:

1. Check if direct substitution works: First, substitute $x = -1$ into the expression:

   $\frac{1}{-1 + 1} - \frac{3}{(-1)^3 + 1} = \frac{1}{0} - \frac{3}{-1 + 1} = \frac{1}{0} - \frac{3}{0}$

   Since this leads to a division by zero (undefined), we cannot directly substitute. This suggests we should look for another method, such as factoring or using limits.

2. Factor the denominators: Let's focus on the second fraction $\frac{3}{x^3 + 1}$. We can factor $x^3 + 1$ using the sum of cubes formula: $x^3 + 1 = (x + 1)(x^2 - x + 1)$ So, the expression becomes: $\frac{1}{x + 1} - \frac{3}{(x + 1)(x^2 - x + 1)}$

3. Combine into a single fraction: Now we can combine the two fractions: [ \frac{1}{