Find all functions $f: R \to R ,g: R \to R$ satisfying the following equality $f(f(x+y))=xf(y)+g(x)$ for all real $x$ and $y$.

I. Gorodnin	
1. Let \( P(x, y) \) be the assertion \( f(f(x+y)) = xf(y) + g(x) \).

2. Consider \( P(0, y) \):
   \[
   f(f(y)) = 0 \cdot f(y) + g(0) \implies f(f(y)) = g(0)
   \]
   This implies that \( f(f(y)) \) is a constant function. Let \( c = g(0) \).

3. Next, consider \( P(1, y) \):
   \[
   f(f(1+y)) = 1 \cdot f(y) + g(1) \implies f(f(1+y)) = f(y) + g(1)
   \]

4. Now, consider \( P(0, x+1) \):
   \[
   f(f(x+1)) = 0 \cdot f(x+1) + g(0) \implies f(f(x+1)) = g(0) = c
   \]

5. Subtracting the equation from step 4 from the equation in step 3:
   \[
   f(y) + g(1) = c
   \]
   This implies:
   \[
   f(y) = c - g(1)
   \]
   Therefore, \( f(y) \) is a constant function. Let \( f(y) = k \) for some constant \( k \).

6. Substitute \( f(y) = k \) back into the original equation:
   \[
   f(f(x+y)) = k = xk + g(x)
   \]
   This simplifies to:
   \[
   k = xk + g(x)
   \]
   Solving for \( g(x) \):
   \[
   g(x) = k(1 - x)
   \]

7. Therefore, the functions \( f \) and \( g \) are:
   \[
   f(x) = k \quad \text{and} \quad g(x) = k(1 - x)
   \]

8. Verify that these functions satisfy the original equation:
   \[
   f(f(x+y)) = k \quad \text{and} \quad xf(y) + g(x) = xk + k(1 - x) = k
   \]
   Both sides are equal, confirming that the functions satisfy the given equation.

Conclusion:
\[
\boxed{(f(x), g(x)) = (k, k(1 - x)) \quad \forall x}
\]