Solve in the real numbers $x, y, z$ a system of the equations:
\[
\begin{cases}
x^2 - (y+z+yz)x + (y+z)yz = 0  \\ 
y^2 - (z + x + zx)y + (z+x)zx = 0 \\
  z^2 - (x+y+xy)z + (x+y)xy = 0. \\
\end{cases}
\]
To solve the given system of equations in real numbers \(x, y, z\):

\[
\begin{cases}
x^2 - (y+z+yz)x + (y+z)yz = 0 \\
y^2 - (z + x + zx)y + (z+x)zx = 0 \\
z^2 - (x+y+xy)z + (x+y)xy = 0
\end{cases}
\]

we will analyze each equation step by step.

1. **First Equation Analysis:**
   \[
   x^2 - (y+z+yz)x + (y+z)yz = 0
   \]
   This is a quadratic equation in \(x\). Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -(y+z+yz)\), and \(c = (y+z)yz\), we get:
   \[
   x = \frac{y+z+yz \pm \sqrt{(y+z+yz)^2 - 4(y+z)yz}}{2}
   \]
   Simplifying the discriminant:
   \[
   (y+z+yz)^2 - 4(y+z)yz = y^2 + 2yz + z^2 + y^2z^2 + 2yz(y+z) - 4yz(y+z)
   \]
   \[
   = y^2 + 2yz + z^2 + y^2z^2 + 2y^2z + 2yz^2 - 4y^2z - 4yz^2
   \]
   \[
   = y^2 + 2yz + z^2 + y^2z^2 - 2y^2z - 2yz^2
   \]
   \[
   = y^2 + z^2 + y^2z^2
   \]
   Therefore, the solutions for \(x\) are:
   \[
   x = y+z \quad \text{or} \quad x = yz
   \]

2. **Second Equation Analysis:**
   \[
   y^2 - (z + x + zx)y + (z+x)zx = 0
   \]
   Similarly, this is a quadratic equation in \(y\). Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -(z + x + zx)\), and \(c = (z+x)zx\), we get:
   \[
   y = \frac{z + x + zx \pm \sqrt{(z + x + zx)^2 - 4(z+x)zx}}{2}
   \]
   Simplifying the discriminant:
   \[
   (z + x + zx)^2 - 4(z+x)zx = z^2 + 2zx + x^2 + z^2x^2 + 2zx(z+x) - 4zx(z+x)
   \]
   \[
   = z^2 + 2zx + x^2 + z^2x^2 + 2z^2x + 2zx^2 - 4z^2x - 4zx^2
   \]
   \[
   = z^2 + x^2 + z^2x^2
   \]
   Therefore, the solutions for \(y\) are:
   \[
   y = z + x \quad \text{or} \quad y = zx
   \]

3. **Third Equation Analysis:**
   \[
   z^2 - (x+y+xy)z + (x+y)xy = 0
   \]
   Similarly, this is a quadratic equation in \(z\). Using the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -(x+y+xy)\), and \(c = (x+y)xy\), we get:
   \[
   z = \frac{x+y+xy \pm \sqrt{(x+y+xy)^2 - 4(x+y)xy}}{2}
   \]
   Simplifying the discriminant:
   \[
   (x+y+xy)^2 - 4(x+y)xy = x^2 + 2xy + y^2 + x^2y^2 + 2xy(x+y) - 4xy(x+y)
   \]
   \[
   = x^2 + 2xy + y^2 + x^2y^2 + 2x^2y + 2xy^2 - 4x^2y - 4xy^2
   \]
   \[
   = x^2 + y^2 + x^2y^2
   \]
   Therefore, the solutions for \(z\) are:
   \[
   z = x + y \quad \text{or} \quad z = xy
   \]

4. **Combining Solutions:**
   From the above analysis, we have the following possible solutions:
   \[
   x = y + z \quad \text{or} \quad x = yz
   \]
   \[
   y = z + x \quad \text{or} \quad y = zx
   \]
   \[
   z = x + y \quad \text{or} \quad z = xy
   \]

   By substituting and solving these equations, we find the solutions:
   \[
   (0,0,0), (1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1)
   \]

   Additionally, we can have:
   \[
   x = y + z, \quad y = xz, \quad z = xy
   \]
   Solving \(x = x(y+z)\) for \(x \neq 0\), we get:
   \[
   x = 1, \quad y = \frac{1}{2}, \quad z = \frac{1}{2}
   \]
   Thus, \((1, \frac{1}{2}, \frac{1}{2})\) and its cyclic permutations are also solutions.

The final answer is \( \boxed{ (0,0,0), (1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1), (1, \frac{1}{2}, \frac{1}{2}), (\frac{1}{2}, 1, \frac{1}{2}), (\frac{1}{2}, \frac{1}{2}, 1) } \)