Let $\mathbb{N}_0$ denote the set of all nonnegative integers. Determine all functions $f:\mathbb{N}_0\to\mathbb{N}_0$ with the following two properties:

[list]
[*] $0\le f(x)\le x^2$ for all $x\in\mathbb{N}_0$

[*] $x-y$ divides $f(x)-f(y)$ for all $x,y\in\mathbb{N}_0$ with $x>y$[/list]
We need to determine all functions \( f: \mathbb{N}_0 \to \mathbb{N}_0 \) that satisfy the following two properties:
1. \( 0 \le f(x) \le x^2 \) for all \( x \in \mathbb{N}_0 \)
2. \( x - y \) divides \( f(x) - f(y) \) for all \( x, y \in \mathbb{N}_0 \) with \( x > y \)

We will consider two cases: when \( f \) is a constant function and when \( f \) is a non-constant function.

### Case 1: \( f \) is a constant function
If \( f(x) = a \) for some constant \( a \in \mathbb{N}_0 \), then substituting \( x = 0 \) into the first condition, we get:
\[ 0 \le f(0) \le 0 \]
This implies \( f(0) = 0 \). Since \( f \) is constant, \( f(x) = 0 \) for all \( x \in \mathbb{N}_0 \).

### Case 2: \( f \) is a non-constant function
Since \( f(0) = 0 \), substituting \( y = 0 \) into the second condition, we get:
\[ x \mid f(x) \]
for all \( x \in \mathbb{N}_0 \). This means \( f(x) = kx \) for some integer \( k \).

#### Subcase 1: \( f(1) = 0 \)
If \( f(1) = 0 \), then substituting \( y = 1 \) into the second condition, we get:
\[ x - 1 \mid f(x) \]
Since \( x \mid f(x) \) and \( (x, x-1) = 1 \), it follows that:
\[ x(x-1) \mid f(x) \]
Thus, \( x^2 - x \le f(x) \). Given \( f(x) \le x^2 \), we have:
\[ x^2 - x \le f(x) \le x^2 \]
Since \( x \mid f(x) \), the only possible values for \( f(x) \) are \( x^2 - x \) and \( x^2 \). However, if \( f(x) = x^2 \), then \( f(1) = 1 \neq 0 \). Therefore, \( f(x) = x^2 - x \) for all \( x \in \mathbb{N}_0 \).

#### Subcase 2: \( f(1) = 1 \)
If \( f(1) = 1 \), then we have:
\[ x \mid f(x) \]
and
\[ x - 1 \mid f(x) - 1 \]
For any \( k \neq 0, 1 \), let \( f(k) = mk \) for some integer \( m \). Then:
\[ k - 1 \mid mk - 1 \]
This implies:
\[ k - 1 \mid (mk - 1) - (mk - m) = m - 1 \]
Thus, \( k - 1 \mid m - 1 \). Given \( f(k) \le k^2 \), we have \( m \le k \). Since \( k - 1 \mid m - 1 \), if \( m - 1 \neq 0 \), then \( k - 1 \le m - 1 \), implying \( k \le m \). Therefore, \( m - 1 = 0 \) or \( m = k \). Hence, \( f(x) = x \) or \( f(x) = x^2 \).

Now, suppose there exist two numbers \( a \) and \( b \) such that \( f(a) = a \) and \( f(b) = b^2 \) (assuming \( a, b \neq 0, 1 \)). If \( a \ge b \), then:
\[ a - b \mid f(a) - f(b) = a - b^2 \]
This implies:
\[ a - b \le a - b^2 \]
Thus, \( b^2 \le b \), which is impossible for \( b > 1 \). Hence, \( a < b \), meaning \( f(x) = x \) for all \( x \le A \) and \( f(x) = x^2 \) for all \( x > A \) for some constant \( A \). However, this function does not satisfy the conditions.

Therefore, the only solutions are:
\[ f(x) = 0 \]
\[ f(x) = x \]
\[ f(x) = x^2 - x \]
\[ f(x) = x^2 \]

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