Given that $x > -1$, we have $x + 1 > 0$.

The function $y = x + \frac{1}{x + 1}$ can be rewritten as $y = x + 1 + \frac{1}{x + 1} - 1$.

Applying the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality), we have:

$y = x + 1 + \frac{1}{x + 1} - 1 \geq 2 \sqrt{(x + 1) \cdot \frac{1}{x + 1}} - 1 = 2 - 1 = 1$.

Equality holds if and only if $x + 1 = \frac{1}{x + 1}$, which occurs when $x = 0$.

Therefore, the minimum value of the function $y = x + \frac{1}{x + 1}$ is $1$, which corresponds to $x = 0$.

Hence, the answer is $\boxed{0}$.

This problem primarily tests one's understanding of the AM-GM Inequality and its applications.