Find all real numbers$ x$ and $y$ such that $$x^2 + y^2 = 2$$
$$\frac{x^2}{2 - y}+\frac{y^2}{2 - x}= 2.$$
1. Given the equations:
   \[
   x^2 + y^2 = 2
   \]
   \[
   \frac{x^2}{2 - y} + \frac{y^2}{2 - x} = 2
   \]

2. Let \( S = x + y \) and \( P = xy \). Using the identity \( x^2 + y^2 = S^2 - 2P \), we substitute into the first equation:
   \[
   S^2 - 2P = 2
   \]

3. For the second equation, we start by clearing the denominators:
   \[
   \frac{x^2}{2 - y} + \frac{y^2}{2 - x} = 2
   \]
   Multiply both sides by \((2 - y)(2 - x)\):
   \[
   x^2(2 - x) + y^2(2 - y) = 2(2 - x)(2 - y)
   \]

4. Expand both sides:
   \[
   2x^2 - x^3 + 2y^2 - y^3 = 4 - 2x - 2y + xy
   \]

5. Substitute \( x^2 + y^2 = 2 \) into the equation:
   \[
   2(2) - (x^3 + y^3) = 4 - 2(x + y) + xy
   \]
   \[
   4 - (x^3 + y^3) = 4 - 2S + P
   \]

6. Using the identity for the sum of cubes \( x^3 + y^3 = (x + y)(x^2 - xy + y^2) \):
   \[
   x^3 + y^3 = S(S^2 - 3P)
   \]
   Substitute this into the equation:
   \[
   4 - S(S^2 - 3P) = 4 - 2S + P
   \]

7. Simplify and rearrange:
   \[
   4 - S^3 + 3SP = 4 - 2S + P
   \]
   \[
   -S^3 + 3SP = -2S + P
   \]
   \[
   S^3 - 3SP + 2S - P = 0
   \]

8. Substitute \( P = \frac{S^2 - 2}{2} \) from the first equation:
   \[
   S^3 - 3S\left(\frac{S^2 - 2}{2}\right) + 2S - \frac{S^2 - 2}{2} = 0
   \]
   \[
   S^3 - \frac{3S^3 - 6S}{2} + 2S - \frac{S^2 - 2}{2} = 0
   \]
   Multiply through by 2 to clear the fractions:
   \[
   2S^3 - 3S^3 + 6S + 4S - S^2 + 2 = 0
   \]
   \[
   -S^3 + 10S - S^2 + 2 = 0
   \]

9. Rearrange and factor:
   \[
   S^3 - 2S^2 + 2S - 4 = 0
   \]
   \[
   (S - 2)(S^2 + 2) = 0
   \]

10. Solve for \( S \):
    \[
    S = 2 \quad \text{or} \quad S^2 + 2 = 0
    \]
    \[
    S = 2 \quad \text{or} \quad S = \pm i\sqrt{2}
    \]

11. For real solutions, \( S = 2 \). Substitute \( S = 2 \) into \( P = \frac{S^2 - 2}{2} \):
    \[
    P = \frac{2^2 - 2}{2} = \frac{4 - 2}{2} = 1
    \]

12. Solve the quadratic equation \( t^2 - St + P = 0 \):
    \[
    t^2 - 2t + 1 = 0
    \]
    \[
    (t - 1)^2 = 0
    \]
    \[
    t = 1
    \]

13. Therefore, \( x = 1 \) and \( y = 1 \).

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