Determine all functions $ f: \mathbb{R} \mapsto \mathbb{R}$ with $ x,y \in \mathbb{R}$ such that

\[ f(x \minus{} f(y)) \equal{} f(x\plus{}y) \plus{} f(y)\]
**
   Assume \( f \) is not identically zero. From the periodicity and additivity, we can infer that \( f \) must be of the form \( f(x) = kx \) for some constant \( k \). Substitute \( f(x) = kx \) into the original equation:
   \[
   f(x - f(y)) = f(x + y) + f(y) \implies k(x - ky) = k(x + y) + ky
   \]
   Simplifying, we get:
   \[
   kx - k^2y = kx + ky + ky \implies -k^2y = 2ky \implies k(k + 2) = 0
   \]
   This gives us two solutions for \( k \):
   \[
   k = 0 \quad \text{or} \quad k = -2
   \]

7. **Verification:**
   - If \( k = 0 \), then \( f(x) = 0 \), which is a trivial solution.
   - If \( k = -2 \), then \( f(x) = -2x \). Verify this:
     \[
     f(x - f(y)) = f(x + y) + f(y) \implies f(x + 2y) = -2(x + y) + (-2y) = -2x - 4y
     \]
     This satisfies the original equation.

Thus, the solutions to the functional equation are \( f(x) = 0 \) and \( f(x) = -2x \).

The final answer is \( \boxed{ f(x) = 0 } \) or \( f(x) = -2x \).