Find all the quaterns $(x,y,z,w)$ of real numbers (not necessarily distinct) that solve the following system of equations:
$$x+y=z^2+w^2+6zw$$
$$x+z=y^2+w^2+6yw$$
$$x+w=y^2+z^2+6yz$$
$$y+z=x^2+w^2+6xw$$
$$y+w=x^2+z^2+6xz$$
$$z+w=x^2+y^2+6xy$$
1. **Initial Assumption:**
   First, consider the case where all variables are equal, i.e., \(x = y = z = w\). Substituting \(x\) for \(y\), \(z\), and \(w\) in the first equation, we get:
   \[
   x + x = x^2 + x^2 + 6x^2 \implies 2x = 8x^2 \implies x(8x - 2) = 0 \implies x = 0 \text{ or } x = \frac{1}{4}
   \]
   Both solutions \(x = 0\) and \(x = \frac{1}{4}\) work. Thus, the solutions \((0, 0, 0, 0)\) and \(\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}\right)\) are valid.

2. **Assuming \(y \neq z\):**
   Subtracting equation (2) from equation (1) and equation (6) from equation (5), we get:
   \[
   (x + y) - (x + z) = (z^2 + w^2 + 6zw) - (y^2 + w^2 + 6yw) \implies y - z = (z - y)(z + y + 6w)
   \]
   \[
   (y + w) - (z + w) = (x^2 + z^2 + 6xz) - (x^2 + y^2 + 6xy) \implies y - z = (z - y)(x + y + 6x)
   \]
   Since \(y \neq z\), we have:
   \[
   1 + y + z + 6w = 0 \implies x = w
   \]

3. **Assuming \(x = w\):**
   Subtracting equation (4) from equation (2) and equation (5) from equation (3), we get:
   \[
   (x + z) - (y + z) = (y^2 + w^2 + 6yw) - (x^2 + w^2 + 6xw) \implies x - y = (y - x)(1 + x + y + 6w)
   \]
   \[
   (x + w) - (y + w) = (y^2 + z^2 + 6yz) - (x^2 + z^2 + 6xz) \implies x - y = (y - x)(1 + x + y + 6z)
   \]
   Since \(x \neq y\), we have:
   \[
   1 + x + y + 6w = 0 \implies z = w
   \]

4. **Assuming \(x = y\):**
   We now have \(x = y\) and \(z = w\). Substituting these into the equations, we get:
   \[
   x + z = 8x^2
   \]
   \[
   2x = z^2 + x^2 + 6zx \implies 2x = z^2 + 6zx + 9x^2 - 8x^2 \implies (z + 3x)^2 = 8x^2 + 2x = z + 3x
   \]
   This implies:
   \[
   z + 3x = 0 \text{ or } z + 3x = 1
   \]

5. **Case 1: \(z + 3x = 0\):**
   \[
   z = -3x
   \]
   Substituting \(z = -3x\) into the equations, we get:
   \[
   -2x = 8x^2 \implies x = 0 \text{ or } x = -\frac{1}{4}
   \]
   Thus, the solutions are \((0, 0, 0, 0)\) and \(\left(-\frac{1}{4}, -\frac{1}{4}, \frac{3}{4}, -\frac{1}{4}\right)\).

6. **Case 2: \(z + 3x = 1\):**
   \[
   z = 1 - 3x
   \]
   Substituting \(z = 1 - 3x\) into the equations, we get:
   \[
   8x^2 + 2x - 1 = 0 \implies (4x - 1)(2x + 1) = 0 \implies x = \frac{1}{4} \text{ or } x = -\frac{1}{2}
   \]
   Thus, the solutions are \(\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}\right)\) and \(\left(-\frac{1}{2}, -\frac{1}{2}, \frac{5}{2}, -\frac{1}{2}\right)\).

7. **Conclusion:**
   The solutions are:
   \[
   (x, y, z, w) = \{(0, 0, 0, 0), \left(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}\right), \left(-\frac{1}{4}, -\frac{1}{4}, \frac{3}{4}, -\frac{1}{4}\right), \left(-\frac{1}{2}, -\frac{1}{2}, \frac{5}{2}, -\frac{1}{2}\right)\}
   \]
   and all cyclic permutations.

The final answer is \( \boxed{ \{(0, 0, 0, 0), \left(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}\right), \left(-\frac{1}{4}, -\frac{1}{4}, \frac{3}{4}, -\frac{1}{4}\right), \left(-\frac{1}{2}, -\frac{1}{2}, \frac{5}{2}, -\frac{1}{2}\right)\} } \) and all cyclic permutations.