Determine all pairs of integer numbers $(x, y)$ such that:
$$\frac{(x - y)^2}{x + y} = x - y + 6$$
1. Start by substituting \( x - y = s \) into the given equation:
   \[
   \frac{(x - y)^2}{x + y} = x - y + 6 \implies \frac{s^2}{x + y} = s + 6
   \]

2. Express \( x + y \) in terms of \( s \) and \( y \):
   \[
   x + y = s + 2y
   \]

3. Substitute \( x + y = s + 2y \) into the equation:
   \[
   \frac{s^2}{s + 2y} = s + 6
   \]

4. Clear the fraction by multiplying both sides by \( s + 2y \):
   \[
   s^2 = (s + 6)(s + 2y)
   \]

5. Expand the right-hand side:
   \[
   s^2 = s^2 + 6s + 2ys + 12y
   \]

6. Simplify by subtracting \( s^2 \) from both sides:
   \[
   0 = 6s + 2ys + 12y
   \]

7. Factor out the common terms:
   \[
   0 = 2y(s + 6) + 6s
   \]

8. Rearrange the equation:
   \[
   2y(s + 6) + 6s = 0
   \]

9. Factor out the common factor of 6:
   \[
   6(y(s + 6) + s) = 0
   \]

10. Simplify the equation:
    \[
    y(s + 6) + s = 0
    \]

11. Rearrange to solve for \( y \):
    \[
    y(s + 6) = -s \implies y = \frac{-s}{s + 6}
    \]

12. Since \( y \) must be an integer, \( \frac{-s}{s + 6} \) must also be an integer. This implies that \( s + 6 \) must be a divisor of \( -s \).

13. Let \( k = s + 6 \). Then \( s = k - 6 \), and we need \( k \) to be a divisor of \( -(k - 6) \):
    \[
    k \mid -(k - 6) \implies k \mid 6
    \]

14. The divisors of 6 are \( \pm 1, \pm 2, \pm 3, \pm 6 \). We will check each case:

    - For \( k = 1 \):
      \[
      s + 6 = 1 \implies s = -5 \implies y = \frac{-(-5)}{-5 + 6} = \frac{5}{1} = 5
      \]
      Thus, \( x - y = -5 \) and \( x + y = -4 \). Solving these, we get:
      \[
      x - 5 = -5 \implies x = 0, \quad 0 + y = -4 \implies y = -4
      \]
      So, \( (x, y) = (0, -4) \).

    - For \( k = -1 \):
      \[
      s + 6 = -1 \implies s = -7 \implies y = \frac{-(-7)}{-7 + 6} = \frac{7}{-1} = -7
      \]
      Thus, \( x - y = -7 \) and \( x + y = -8 \). Solving these, we get:
      \[
      x + 7 = -7 \implies x = 0, \quad 0 - y = -8 \implies y = 8
      \]
      So, \( (x, y) = (0, 8) \).

    - For \( k = 2 \):
      \[
      s + 6 = 2 \implies s = -4 \implies y = \frac{-(-4)}{-4 + 6} = \frac{4}{2} = 2
      \]
      Thus, \( x - y = -4 \) and \( x + y = -2 \). Solving these, we get:
      \[
      x - 2 = -4 \implies x = -2, \quad -2 + y = -2 \implies y = 0
      \]
      So, \( (x, y) = (-2, 0) \).

    - For \( k = -2 \):
      \[
      s + 6 = -2 \implies s = -8 \implies y = \frac{-(-8)}{-8 + 6} = \frac{8}{-2} = -4
      \]
      Thus, \( x - y = -8 \) and \( x + y = -12 \). Solving these, we get:
      \[
      x + 4 = -8 \implies x = -4, \quad -4 - y = -12 \implies y = 8
      \]
      So, \( (x, y) = (-4, 8) \).

    - For \( k = 3 \):
      \[
      s + 6 = 3 \implies s = -3 \implies y = \frac{-(-3)}{-3 + 6} = \frac{3}{3} = 1
      \]
      Thus, \( x - y = -3 \) and \( x + y = -2 \). Solving these, we get:
      \[
      x - 1 = -3 \implies x = -2, \quad -2 + y = -2 \implies y = 0
      \]
      So, \( (x, y) = (-2, 0) \).

    - For \( k = -3 \):
      \[
      s + 6 = -3 \implies s = -9 \implies y = \frac{-(-9)}{-9 + 6} = \frac{9}{-3} = -3
      \]
      Thus, \( x - y = -9 \) and \( x + y = -12 \). Solving these, we get:
      \[
      x + 3 = -9 \implies x = -6, \quad -6 - y = -12 \implies y = 6
      \]
      So, \( (x, y) = (-6, 6) \).

    - For \( k = 6 \):
      \[
      s + 6 = 6 \implies s = 0 \implies y = \frac{-0}{0 + 6} = 0
      \]
      Thus, \( x - y = 0 \) and \( x + y = 0 \). Solving these, we get:
      \[
      x = 0, \quad y = 0
      \]
      So, \( (x, y) = (0, 0) \).

    - For \( k = -6 \):
      \[
      s + 6 = -6 \implies s = -12 \implies y = \frac{-(-12)}{-12 + 6} = \frac{12}{-6} = -2
      \]
      Thus, \( x - y = -12 \) and \( x + y = -14 \). Solving these, we get:
      \[
      x + 2 = -12 \implies x = -10, \quad -10 - y = -14 \implies y = 4
      \]
      So, \( (x, y) = (-10, 4) \).

The final answer is \( \boxed{ (0, -4), (0, 8), (-2, 0), (-4, 8), (-2, 0), (-6, 6), (0, 0), (-10, 4) } \)