Find all functions $f:\mathbb{N}_0\to\mathbb{N}_0$ for which $f(0)=0$ and
\[f(x^2-y^2)=f(x)f(y) \]
for all $x,y\in\mathbb{N}_0$ with $x>y$.
1. **Initial Condition and Simplification:**
   Given \( f(0) = 0 \) and the functional equation \( f(x^2 - y^2) = f(x)f(y) \) for all \( x, y \in \mathbb{N}_0 \) with \( x > y \).

2. **Substitute \( y = 0 \):**
   \[
   f(x^2 - 0^2) = f(x)f(0) \implies f(x^2) = f(x) \cdot 0 \implies f(x^2) = 0
   \]
   This implies that \( f(x^2) = 0 \) for all \( x \in \mathbb{N}_0 \).

3. **Substitute \( x = y + 1 \):**
   \[
   f((y+1)^2 - y^2) = f(y+1)f(y) \implies f(2y + 1) = f(y+1)f(y)
   \]
   Since \( f(x^2) = 0 \) for all \( x \), we have \( f(y+1)^2 = 0 \) which implies \( f(y+1) = 0 \) for all \( y \in \mathbb{N}_0 \).

4. **Substitute \( x = y + 2 \):**
   \[
   f((y+2)^2 - y^2) = f(y+2)f(y) \implies f(4y + 4) = f(y+2)f(y)
   \]
   Since \( f(x^2) = 0 \) for all \( x \), we have \( f(y+2)f(y) = 0 \). Given that \( f(y) = 0 \) for all \( y \), this equation holds true.

5. **General Case for \( 4k + 2 \):**
   For \( x = 2k + 1 \) and \( y = 2k \):
   \[
   f((2k+1)^2 - (2k)^2) = f(2k+1)f(2k) \implies f(4k + 1) = f(2k+1)f(2k)
   \]
   Since \( f(2k+1) = 0 \) and \( f(2k) = 0 \), we have \( f(4k + 1) = 0 \).

6. **Conclusion:**
   From the above steps, we see that \( f(x) = 0 \) for all \( x \in \mathbb{N}_0 \). The only function that satisfies the given conditions is \( f(x) = 0 \) for all \( x \).

The final answer is \( \boxed{ f(x) = 0 } \) for all \( x \in \mathbb{N}_0 \).