Find all monotonic functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying
\[f(f(x) - y) + f(x+y) = 0,\]
for every real $x, y$.
(Note that monotonic means that function is not increasing or not decreasing)
1. Let \( P(x, y) \) be the assertion \( f(f(x) - y) + f(x + y) = 0 \).

2. **Case 1: \( f(x) \) is not injective.**
   - Assume there exist \( a \) and \( b \) such that \( a \neq b \) and \( f(a) = f(b) \).
   - Consider \( P(a, x - b) \) and \( P(b, x - b) \):
     \[
     P(a, x - b): f(f(a) - (x - b)) + f(a + (x - b)) = 0 \implies f(f(a) - x + b) + f(a + x - b) = 0
     \]
     \[
     P(b, x - b): f(f(b) - (x - b)) + f(b + (x - b)) = 0 \implies f(f(b) - x + b) + f(b + x - b) = 0
     \]
   - Since \( f(a) = f(b) \), we have:
     \[
     f(f(a) - x + b) + f(a + x - b) = 0 \quad \text{and} \quad f(f(a) - x + b) + f(b + x - b) = 0
     \]
   - This implies:
     \[
     f(a + x - b) = f(b + x - b) \implies f(x + a - b) = f(x)
     \]
   - Therefore, \( f(x) \) is periodic with period \( a - b \). Since \( f(x) \) is monotonic, it must be constant.
   - Let \( f(x) = c \). Substituting into the original equation:
     \[
     f(f(x) - y) + f(x + y) = 0 \implies f(c - y) + f(x + y) = 0
     \]
     \[
     c + c = 0 \implies c = 0
     \]
   - Hence, \( f(x) = 0 \) for all \( x \).

3. **Case 2: \( f(x) \) is injective.**
   - Consider \( P(x, -f(y)) \):
     \[
     f(f(x) + f(y)) + f(x - f(y)) = 0
     \]
   - Consider \( P(y, -f(x)) \):
     \[
     f(f(y) + f(x)) + f(y - f(x)) = 0
     \]
   - From the above two equations, we have:
     \[
     f(f(x) + f(y)) + f(x - f(y)) = 0 \quad \text{and} \quad f(f(x) + f(y)) + f(y - f(x)) = 0
     \]
   - This implies:
     \[
     f(x - f(y)) = f(y - f(x))
     \]
   - Since \( f \) is injective, we have:
     \[
     x - f(y) = y - f(x) \implies f(x) + x = f(y) + y
     \]
   - Let \( f(x) + x = c \) for some constant \( c \). Then:
     \[
     f(x) = c - x
     \]
   - Substituting \( f(x) = c - x \) into the original equation:
     \[
     f(f(x) - y) + f(x + y) = 0 \implies f(c - x - y) + f(x + y) = 0
     \]
     \[
     f(c - x - y) = c - (c - x - y) = x + y
     \]
     \[
     f(x + y) = c - (x + y)
     \]
     \[
     c - (x + y) + c - (x + y) = 0 \implies 2c - 2(x + y) = 0 \implies c = 0
     \]
   - Hence, \( f(x) = -x \).

4. Therefore, the only two solutions are:
   - \( f(x) = 0 \) for all \( x \)
   - \( f(x) = -x \) for all \( x \)

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