Let $a,b$ be two positive integers, such that $ab\neq 1$. Find all the integer values that $f(a,b)$ can take, where \[ f(a,b) = \frac { a^2+ab+b^2} { ab- 1} . \]
1. **Initial Observations and Specific Cases**:
   - We are given \( f(a,b) = \frac{a^2 + ab + b^2}{ab - 1} \).
   - We need to find all possible integer values of \( f(a,b) \) for positive integers \( a \) and \( b \) such that \( ab \neq 1 \).

2. **Checking Specific Values**:
   - For \( a = 2 \) and \( b = 1 \):
     \[
     f(2,1) = \frac{2^2 + 2 \cdot 1 + 1^2}{2 \cdot 1 - 1} = \frac{4 + 2 + 1}{2 - 1} = \frac{7}{1} = 7
     \]
   - For \( a = 2 \) and \( b = 2 \):
     \[
     f(2,2) = \frac{2^2 + 2 \cdot 2 + 2^2}{2 \cdot 2 - 1} = \frac{4 + 4 + 4}{4 - 1} = \frac{12}{3} = 4
     \]

3. **General Case for \( a = b \)**:
   - If \( a = b \), then:
     \[
     f(a,a) = \frac{a^2 + a \cdot a + a^2}{a \cdot a - 1} = \frac{3a^2}{a^2 - 1} = 3 + \frac{3}{a^2 - 1}
     \]
   - For \( f(a,a) \) to be an integer, \( \frac{3}{a^2 - 1} \) must be an integer. The only positive integer \( a \) that satisfies this is \( a = 2 \), giving:
     \[
     f(2,2) = 4
     \]

4. **Considering \( \min\{a,b\} \leq 2 \)**:
   - If \( \min\{a,b\} \leq 2 \), we have already checked the cases \( (2,1) \) and \( (2,2) \), which yield \( f(a,b) \in \{4,7\} \).

5. **Assuming \( k \neq 4,7 \) and Minimality Argument**:
   - Suppose \( k \neq 4,7 \) is in the range of \( f \), and pick \( (a,b) \) such that \( f(a,b) = k \), \( a > b \), and \( a + b \) is minimal. Assume \( b > 2 \).
   - Rewrite \( f(a,b) = k \) as a quadratic in \( a \):
     \[
     a^2 - (k-1)ba + (b^2 + k) = 0
     \]
   - By Vieta's formulas, the other root is \( \frac{b^2 + k}{a} \). By minimality, we must have:
     \[
     \frac{b^2 + k}{a} \geq a \implies k \geq a^2 - b^2 \geq a^2 - (a-1)^2 = 2a - 1
     \]

6. **Deriving a Contradiction**:
   - This implies:
     \[
     k \geq 2a - 1
     \]
   - Substituting \( k = \frac{a^2 + ab + b^2}{ab - 1} \) and simplifying:
     \[
     \frac{a^2 + ab + b^2}{ab - 1} \geq 2a - 1
     \]
     \[
     a^2 + ab + b^2 \geq (2a - 1)(ab - 1)
     \]
     \[
     a^2 + ab + b^2 \geq 2a^2b - 2a - ab + 1
     \]
     \[
     3a^2 + ab + b^2 \geq 2a^2b - 2a + 1
     \]
     \[
     3a^2 \geq 2a^2b - ab - 2a + 1
     \]
     \[
     3a^2 > a^2b
     \]
   - This inequality only holds for \( b \leq 2 \), which contradicts our assumption that \( b > 2 \).

7. **Conclusion**:
   - Therefore, such a \( k \) does not exist, and the only possible integer values for \( f(a,b) \) are \( 4 \) and \( 7 \).

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