Determine all the possible non-negative integer values that are able to satisfy the expression:
$\frac{(m^2+mn+n^2)}{(mn-1)}$ 
if $m$ and $n$ are non-negative integers such that $mn \neq 1$.
To determine all possible non-negative integer values that satisfy the expression $\frac{m^2 + mn + n^2}{mn - 1}$, given that $m$ and $n$ are non-negative integers and $mn \neq 1$, we will analyze the expression step-by-step.

1. **Case 1: \( m = 0 \) or \( n = 0 \)**
   - If \( m = 0 \), then the expression becomes:
     \[
     \frac{0^2 + 0 \cdot n + n^2}{0 \cdot n - 1} = \frac{n^2}{-1} = -n^2
     \]
     Since \( n \) is a non-negative integer, \( -n^2 \) is a non-positive integer. The only non-negative integer value for \( n \) that makes \( -n^2 \) an integer is \( n = 0 \). Thus, \( (m, n) = (0, 0) \) is a solution, and the expression evaluates to 0.
   - If \( n = 0 \), a similar argument shows that \( (m, n) = (0, 0) \) is a solution, and the expression evaluates to 0.

2. **Case 2: \( m, n > 0 \)**
   - Let \( N = \frac{m^2 + mn + n^2}{mn - 1} \) be an integer.
   - If \( m = n \), then:
     \[
     N = \frac{3m^2}{m^2 - 1} = 3 + \frac{3}{m^2 - 1}
     \]
     For \( N \) to be an integer, \( \frac{3}{m^2 - 1} \) must be an integer. This implies \( m^2 - 1 \) must be a divisor of 3. The possible values for \( m \) are 2 and 4, giving:
     \[
     m = 2 \implies N = 3 + \frac{3}{3} = 4
     \]
     \[
     m = 4 \implies N = 3 + \frac{3}{15} \quad \text{(not an integer)}
     \]
     Thus, \( (m, n) = (2, 2) \) is a solution, and \( N = 4 \).

3. **General Case: \( m \neq n \)**
   - Assume \( m > n > 0 \). We need to show that \( N \) can take specific values.
   - Consider the quadratic equation derived from the expression:
     \[
     x^2 - (N-1)yx + y^2 + N = 0
     \]
     This equation must have integer solutions for \( x \) and \( y \). By analyzing the discriminant and possible values, we can derive constraints on \( m \) and \( n \).

4. **Checking specific values for \( n \)**
   - For \( n = 1 \):
     \[
     N = m + 2 + \frac{3}{m - 1}
     \]
     For \( N \) to be an integer, \( \frac{3}{m - 1} \) must be an integer. Possible values for \( m \) are 2 and 4, giving:
     \[
     m = 2 \implies N = 2 + 2 + \frac{3}{1} = 7
     \]
     \[
     m = 4 \implies N = 4 + 2 + \frac{3}{3} = 7
     \]
     Thus, \( (m, n) = (2, 1) \) and \( (4, 1) \) are solutions, and \( N = 7 \).

   - For \( n = 2 \):
     \[
     4N = 2m + 5 + \frac{21}{2m - 1}
     \]
     Solving for integer values, we find:
     \[
     m = 4 \implies N = 4
     \]
     \[
     m = 11 \implies N = 7
     \]
     Thus, \( (m, n) = (4, 2) \) and \( (11, 2) \) are solutions, and \( N = 4 \) or \( 7 \).

   - For \( n = 3 \):
     \[
     9N = 3m + 10 + \frac{91}{3m - 1}
     \]
     Solving for integer values, we find no solutions.

Thus, the only possible values for \( N \) are 0, 4, and 7.

The final answer is \( \boxed{0, 4, 7} \).