Determine all functions $f : N_0 \to R$ satisfying $f (x+y)+ f (x-y)= f (3x)$ for all $x,y$.
1. **Assertion and Initial Conditions:**
   Let \( P(x, y) \) be the assertion \( f(x+y) + f(x-y) = f(3x) \).

2. **Evaluate at \( P(0, 0) \):**
   \[
   P(0, 0) \implies f(0+0) + f(0-0) = f(3 \cdot 0) \implies f(0) + f(0) = f(0) \implies 2f(0) = f(0) \implies f(0) = 0
   \]

3. **Evaluate at \( P(x, 0) \):**
   \[
   P(x, 0) \implies f(x+0) + f(x-0) = f(3x) \implies f(x) + f(x) = f(3x) \implies 2f(x) = f(3x)
   \]

4. **Evaluate at \( P(x, x) \):**
   \[
   P(x, x) \implies f(x+x) + f(x-x) = f(3x) \implies f(2x) + f(0) = f(3x) \implies f(2x) = f(3x)
   \]
   From step 3, we have \( f(3x) = 2f(x) \). Thus,
   \[
   f(2x) = 2f(x)
   \]

5. **Evaluate at \( P(x+y, x-y) \):**
   \[
   P(x+y, x-y) \implies f((x+y) + (x-y)) + f((x+y) - (x-y)) = f(3(x+y)) \implies f(2x) + f(2y) = f(3(x+y))
   \]
   Using \( f(2x) = 2f(x) \) and \( f(2y) = 2f(y) \), we get:
   \[
   2f(x) + 2f(y) = f(3(x+y))
   \]
   From step 3, we know \( f(3(x+y)) = 2f(x+y) \). Thus,
   \[
   2f(x) + 2f(y) = 2f(x+y) \implies f(x) + f(y) = f(x+y)
   \]
   This is Cauchy's functional equation.

6. **Solution to Cauchy's Functional Equation:**
   Since \( f : \mathbb{N}_0 \to \mathbb{R} \) and \( f(x) + f(y) = f(x+y) \), the only solution is of the form \( f(x) = cx \) for some constant \( c \).

7. **Determine the Constant \( c \):**
   Substitute \( f(x) = cx \) into the original equation:
   \[
   f(x+y) + f(x-y) = f(3x) \implies c(x+y) + c(x-y) = c(3x) \implies cx + cy + cx - cy = 3cx \implies 2cx = 3cx
   \]
   This implies:
   \[
   cx = 0 \implies c = 0
   \]
   Therefore, \( f(x) = 0 \) for all \( x \in \mathbb{N}_0 \).

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