Find all real solutions of the equation: $$x=\frac{2z^2}{1+z^2}$$ $$y=\frac{2x^2}{1+x^2}$$ $$z=\frac{2y^2}{1+y^2}$$
1. We start with the given system of equations:
   \[
   x = \frac{2z^2}{1+z^2}
   \]
   \[
   y = \frac{2x^2}{1+x^2}
   \]
   \[
   z = \frac{2y^2}{1+y^2}
   \]

2. We need to find all real solutions \((x, y, z)\) that satisfy these equations. First, observe that the function \( f(a) = \frac{2a^2}{1+a^2} \) is bounded by 1 for all real \(a\):
   \[
   f(a) = \frac{2a^2}{1+a^2} \leq 1
   \]
   This is because:
   \[
   \frac{2a^2}{1+a^2} \leq 1 \implies 2a^2 \leq 1 + a^2 \implies a^2 \leq 1 \implies a \leq 1
   \]

3. Given the symmetry and the bounded nature of the function, we can infer that:
   \[
   x = \frac{2z^2}{1+z^2} \leq z
   \]
   \[
   y = \frac{2x^2}{1+x^2} \leq x
   \]
   \[
   z = \frac{2y^2}{1+y^2} \leq y
   \]

4. Combining these inequalities, we get:
   \[
   x \leq z \leq y \leq x
   \]
   This implies:
   \[
   x = y = z
   \]

5. Substituting \(x = y = z\) into any of the original equations, we get:
   \[
   x = \frac{2x^2}{1+x^2}
   \]

6. Solving this equation for \(x\):
   \[
   x(1 + x^2) = 2x^2
   \]
   \[
   x + x^3 = 2x^2
   \]
   \[
   x^3 - 2x^2 + x = 0
   \]
   \[
   x(x^2 - 2x + 1) = 0
   \]
   \[
   x(x - 1)^2 = 0
   \]

7. The solutions to this equation are:
   \[
   x = 0 \quad \text{or} \quad x = 1
   \]

8. Therefore, the solutions to the original system are:
   \[
   (x, y, z) = (0, 0, 0) \quad \text{or} \quad (1, 1, 1)
   \]

The final answer is \( \boxed{ (0, 0, 0) } \) or \((1, 1, 1)\)