Find all positive real numbers $(x,y,z)$ such that:

$$x = \frac{1}{y^2+y-1}$$
$$y = \frac{1}{z^2+z-1}$$
$$z = \frac{1}{x^2+x-1}$$
Given the system of equations:
\[ x = \frac{1}{y^2 + y - 1} \]
\[ y = \frac{1}{z^2 + z - 1} \]
\[ z = \frac{1}{x^2 + x - 1} \]

We need to find all positive real numbers \((x, y, z)\) that satisfy these equations.

1. **Rewrite the equations:**
   \[ y(y^2 + y - 1) = 1 \implies y^3 + y^2 - y - 1 = 0 \]
   \[ z(z^2 + z - 1) = 1 \implies z^3 + z^2 - z - 1 = 0 \]
   \[ x(x^2 + x - 1) = 1 \implies x^3 + x^2 - x - 1 = 0 \]

2. **Assume \(x = y = z\):**
   If \(x = y = z\), then:
   \[ x = \frac{1}{x^2 + x - 1} \]
   Multiplying both sides by \(x^2 + x - 1\):
   \[ x(x^2 + x - 1) = 1 \]
   \[ x^3 + x^2 - x - 1 = 0 \]

3. **Solve the cubic equation:**
   The cubic equation \(x^3 + x^2 - x - 1 = 0\) can be solved using the Rational Root Theorem or numerical methods. However, we can also check for simple roots:
   \[ x = 1 \]
   Substituting \(x = 1\):
   \[ 1^3 + 1^2 - 1 - 1 = 0 \]
   \[ 1 + 1 - 1 - 1 = 0 \]
   \[ 0 = 0 \]
   Thus, \(x = 1\) is a solution.

4. **Check for other possible solutions:**
   We need to check if there are other roots of the cubic equation \(x^3 + x^2 - x - 1 = 0\). Using numerical methods or factoring, we find that the other roots are:
   \[ x = 2\cos\left(\frac{2\pi}{9}\right), \quad x = 2\cos\left(\frac{4\pi}{9}\right), \quad x = 2\cos\left(\frac{8\pi}{9}\right) \]

5. **Verify the solutions:**
   For each of these roots, we need to verify if they satisfy the original system of equations. Since the equations are symmetric and cyclic, if \(x = y = z\) is a solution, then it will satisfy all three equations.

6. **Conclusion:**
   The solutions to the system are:
   \[ x = y = z = 1 \]
   \[ x = y = z = 2\cos\left(\frac{2\pi}{9}\right) \]
   \[ x = y = z = 2\cos\left(\frac{4\pi}{9}\right) \]
   \[ x = y = z = 2\cos\left(\frac{8\pi}{9}\right) \]

The final answer is \( \boxed{ (x, y, z) = (1, 1, 1) } \), \((2\cos\left(\frac{2\pi}{9}\right), 2\cos\left(\frac{2\pi}{9}\right), 2\cos\left(\frac{2\pi}{9}\right))\), \((2\cos\left(\frac{4\pi}{9}\right), 2\cos\left(\frac{4\pi}{9}\right), 2\cos\left(\frac{4\pi}{9}\right))\), \((2\cos\left(\frac{8\pi}{9}\right), 2\cos\left(\frac{8\pi}{9}\right), 2\cos\left(\frac{8\pi}{9}\right))\).