Determine all numbers $x, y$ and $z$ satisfying the system of equations
$$\begin{cases} x^2 + yz = 1 \\ y^2 - xz = 0 \\ z^2 + xy = 1\end{cases}$$
To solve the system of equations:
\[
\begin{cases}
x^2 + yz = 1 \\
y^2 - xz = 0 \\
z^2 + xy = 1
\end{cases}
\]

1. **Subtract the first and third equations:**
   \[
   x^2 + yz - (z^2 + xy) = 1 - 1
   \]
   Simplifying, we get:
   \[
   x^2 + yz - z^2 - xy = 0 \implies x^2 - z^2 + yz - xy = 0
   \]
   Factoring, we have:
   \[
   (x - z)(x + z) + y(z - x) = 0 \implies (x - z)(x + z - y) = 0
   \]
   This implies:
   \[
   x = z \quad \text{or} \quad y = x + z
   \]

2. **Case 1: \( x = z \)**
   - Substitute \( z = x \) into the original system:
     \[
     \begin{cases}
     x^2 + yx = 1 \\
     y^2 - x^2 = 0 \\
     x^2 + xy = 1
     \end{cases}
     \]
   - The second equation simplifies to:
     \[
     y^2 = x^2 \implies y = \pm x
     \]
   - Substitute \( y = x \) into the first equation:
     \[
     x^2 + x^2 = 1 \implies 2x^2 = 1 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{\sqrt{2}}{2}
     \]
   - Therefore, \( x = y = z = \pm \frac{\sqrt{2}}{2} \).

3. **Case 2: \( y = x + z \)**
   - Substitute \( y = x + z \) into the original system:
     \[
     \begin{cases}
     x^2 + z(x + z) = 1 \\
     (x + z)^2 - xz = 0 \\
     z^2 + x(x + z) = 1
     \end{cases}
     \]
   - Simplify the first equation:
     \[
     x^2 + zx + z^2 = 1
     \]
   - Simplify the second equation:
     \[
     x^2 + 2xz + z^2 - xz = 0 \implies x^2 + xz + z^2 = 0
     \]
   - Compare the simplified first and second equations:
     \[
     x^2 + zx + z^2 = 1 \quad \text{and} \quad x^2 + xz + z^2 = 0
     \]
     This is a contradiction since \( 1 \neq 0 \).

Therefore, the only valid solution is from Case 1.

The final answer is \( \boxed{ x = y = z = \pm \frac{\sqrt{2}}{2} } \).