Let $n \geq 2$ be a given integer. We aim to find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$f(x - f(y)) = f(x + y^n) + f(f(y) + y^n), \quad \forall x, y \in \mathbb{R}.$$

The solutions to this functional equation are:
1. $f(x) = 0$ for all $x \in \mathbb{R}$.
2. $f(x) = -x^n$ for all $x \in \mathbb{R}$.

To verify, we check both functions:

1. For $f(x) = 0$:
$$f(x - f(y)) = f(x - 0) = 0,$$
$$f(x + y^n) + f(f(y) + y^n) = 0 + 0 = 0,$$
which satisfies the given equation.

2. For $f(x) = -x^n$:
$$f(x - f(y)) = f(x + y^n) = -(x + y^n)^n,$$
$$f(x + y^n) + f(f(y) + y^n) = - (x + y^n)^n + (-(-y^n + y^n)^n) = -(x + y^n)^n,$$
which also satisfies the given equation.

Thus, the only solutions are $f(x) = 0$ and $f(x) = -x^n$.

The answer is: $\boxed{f(x)}$ = 0 $\text{}$ or } f(x) = -x^n}.