Find all pairs of real numbers $(x, y)$, that satisfy the system of equations: $$\left\{\begin{matrix} 6(1-x)^2=\frac{1}{y},\\6(1-y)^2=\frac{1}{x}.\end{matrix}\right.$$
To find all pairs of real numbers \((x, y)\) that satisfy the system of equations:
\[
\left\{
\begin{matrix}
6(1-x)^2 = \frac{1}{y}, \\
6(1-y)^2 = \frac{1}{x}
\end{matrix}
\right.
\]

1. Start by rewriting the given equations:
   \[
   6(1-x)^2 = \frac{1}{y} \quad \text{and} \quad 6(1-y)^2 = \frac{1}{x}
   \]

2. Multiply both sides of each equation by \(y\) and \(x\) respectively:
   \[
   6y(1-x)^2 = 1 \quad \text{and} \quad 6x(1-y)^2 = 1
   \]

3. Equate the two expressions:
   \[
   6y(1-x)^2 = 6x(1-y)^2
   \]

4. Simplify the equation:
   \[
   y(1-x)^2 = x(1-y)^2
   \]

5. Expand both sides:
   \[
   y(1 - 2x + x^2) = x(1 - 2y + y^2)
   \]

6. Rearrange the terms:
   \[
   y - 2xy + yx^2 = x - 2xy + xy^2
   \]

7. Combine like terms:
   \[
   y + yx^2 = x + xy^2
   \]

8. Rearrange to isolate terms involving \(x\) and \(y\):
   \[
   yx^2 - xy^2 = x - y
   \]

9. Factor out common terms:
   \[
   xy(x - y) = x - y
   \]

10. Since \(x - y \neq 0\) (otherwise, the terms would cancel out), divide both sides by \(x - y\):
    \[
    xy = 1
    \]

11. Substitute \(y = \frac{1}{x}\) into the first equation:
    \[
    6(1-x)^2 = x
    \]

12. Solve the quadratic equation:
    \[
    6(1 - 2x + x^2) = x \implies 6x^2 - 12x + 6 = x \implies 6x^2 - 13x + 6 = 0
    \]

13. Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 6\), \(b = -13\), and \(c = 6\):
    \[
    x = \frac{13 \pm \sqrt{169 - 144}}{12} = \frac{13 \pm 5}{12}
    \]

14. Solve for \(x\):
    \[
    x = \frac{18}{12} = \frac{3}{2} \quad \text{or} \quad x = \frac{8}{12} = \frac{2}{3}
    \]

15. Corresponding \(y\) values are:
    \[
    y = \frac{1}{x} \implies y = \frac{2}{3} \quad \text{or} \quad y = \frac{3}{2}
    \]

16. Therefore, the pairs \((x, y)\) are:
    \[
    \left(\frac{3}{2}, \frac{2}{3}\right) \quad \text{and} \quad \left(\frac{2}{3}, \frac{3}{2}\right)
    \]

17. Consider the case \(x = y\):
    \[
    6(1-x)^2 = \frac{1}{x}
    \]

18. Substitute \(x = y\) into the equation:
    \[
    6(1-x)^2 = \frac{1}{x} \implies 6x^3 - 12x^2 + 6x - 1 = 0
    \]

19. Solve the cubic equation using the cubic formula or numerical methods:
    \[
    x = \frac{1}{3}(2 + \sqrt[3]{2} + \frac{1}{\sqrt[3]{2}})
    \]

20. Therefore, the pair \((x, y)\) is:
    \[
    \left(\frac{1}{3}(2 + \sqrt[3]{2} + \frac{1}{\sqrt[3]{2}}), \frac{1}{3}(2 + \sqrt[3]{2} + \frac{1}{\sqrt[3]{2}})\right)
    \]

The final answer is \( \boxed{ \left(\frac{3}{2}, \frac{2}{3}\right) } \), \(\left(\frac{2}{3}, \frac{3}{2}\right)\), and \(\left(\frac{1}{3}(2 + \sqrt[3]{2} + \frac{1}{\sqrt[3]{2}}), \frac{1}{3}(2 + \sqrt[3]{2} + \frac{1}{\sqrt[3]{2}})\right)\).