Find all pairs of integers $(x,y)$ satisfying the following equality $8x^2y^2 + x^2 + y^2 = 10xy$
1. Start with the given equation:
   \[
   8x^2y^2 + x^2 + y^2 = 10xy
   \]

2. Introduce the symmetric sums \( S = x + y \) and \( P = xy \). Using these, we can express \( x^2 + y^2 \) and \( x^2y^2 \) in terms of \( S \) and \( P \):
   \[
   x^2 + y^2 = S^2 - 2P
   \]
   \[
   x^2y^2 = P^2
   \]

3. Substitute these expressions into the original equation:
   \[
   8P^2 + (S^2 - 2P) = 10P
   \]

4. Simplify the equation:
   \[
   8P^2 + S^2 - 2P = 10P
   \]
   \[
   8P^2 + S^2 - 12P = 0
   \]

5. Solve for \( S^2 \):
   \[
   S^2 = 12P - 8P^2
   \]
   \[
   S^2 = 4P(3 - 2P)
   \]

6. For \( S \) to be a real number, the expression inside the square root must be non-negative:
   \[
   4P(3 - 2P) \geq 0
   \]

7. Solve the inequality:
   \[
   P(3 - 2P) \geq 0
   \]
   The roots of the quadratic equation \( P(3 - 2P) = 0 \) are \( P = 0 \) and \( P = \frac{3}{2} \). The quadratic opens downwards, so the inequality holds for:
   \[
   0 \leq P \leq \frac{3}{2}
   \]

8. Since \( S \) and \( P \) must be integers, we check the integer values within this range:
   \[
   P \in \{0, 1\}
   \]

9. For each value of \( P \), solve for \( S \):
   - If \( P = 0 \):
     \[
     S^2 = 4 \cdot 0 \cdot (3 - 2 \cdot 0) = 0
     \]
     \[
     S = 0
     \]
     This corresponds to the pair \((x, y)\) such that \( x + y = 0 \) and \( xy = 0 \). The integer solutions are:
     \[
     (x, y) = (0, 0)
     \]

   - If \( P = 1 \):
     \[
     S^2 = 4 \cdot 1 \cdot (3 - 2 \cdot 1) = 4
     \]
     \[
     S = \pm 2
     \]
     This corresponds to the pairs \((x, y)\) such that \( x + y = 2 \) or \( x + y = -2 \) and \( xy = 1 \). The integer solutions are:
     \[
     (x, y) = (1, 1) \quad \text{or} \quad (x, y) = (-1, -1)
     \]

10. Combine all solutions:
    \[
    (x, y) \in \{(0, 0), (1, 1), (-1, -1)\}
    \]

The final answer is \( \boxed{ \{(0, 0), (1, 1), (-1, -1)\} } \)