Find all the triplets of real numbers $(x , y , z)$ such that :

$y=\frac{x^3+12x}{3x^2+4}$ , $z=\frac{y^3+12y}{3y^2+4}$ , $x=\frac{z^3+12z}{3z^2+4}$
1. We start with the given equations:
   \[
   y = \frac{x^3 + 12x}{3x^2 + 4}
   \]
   \[
   z = \frac{y^3 + 12y}{3y^2 + 4}
   \]
   \[
   x = \frac{z^3 + 12z}{3z^2 + 4}
   \]

2. Assume without loss of generality (W.L.O.G.) that \( x \geq y \geq z \). This implies:
   \[
   x \geq y \geq z
   \]
   Substituting the expressions for \( y \) and \( z \) in terms of \( x \), we get:
   \[
   \frac{z^3 + 12z}{3z^2 + 4} \geq \frac{x^3 + 12x}{3x^2 + 4} \geq \frac{y^3 + 12y}{3y^2 + 4}
   \]
   Since \( x \geq y \geq z \), we also have:
   \[
   z \geq x \geq y
   \]
   This implies \( x = y = z \).

3. Substitute \( x = y = z \) into the first equation:
   \[
   x = \frac{x^3 + 12x}{3x^2 + 4}
   \]

4. Simplify the equation:
   \[
   x(3x^2 + 4) = x^3 + 12x
   \]
   \[
   3x^3 + 4x = x^3 + 12x
   \]
   \[
   2x^3 - 8x = 0
   \]
   \[
   2x(x^2 - 4) = 0
   \]
   \[
   2x(x - 2)(x + 2) = 0
   \]

5. Solve for \( x \):
   \[
   x = 0, \quad x = 2, \quad x = -2
   \]

6. Since \( x = y = z \), the triplets are:
   \[
   (x, y, z) = (0, 0, 0), \quad (2, 2, 2), \quad (-2, -2, -2)
   \]

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