Find all triplets $ (x,y,z) $ of real numbers such that
\[ 2x^3 + 1 = 3zx \]\[ 2y^3 + 1 = 3xy \]\[ 2z^3 + 1 =  3yz \]
We are given the system of equations:
\[ 2x^3 + 1 = 3zx \]
\[ 2y^3 + 1 = 3xy \]
\[ 2z^3 + 1 = 3yz \]

We need to find all triplets \((x, y, z)\) of real numbers that satisfy these equations. We will consider different cases based on the signs of \(x\), \(y\), and \(z\).

### Case 1: \(x, y, z \ge 0\)
Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
\[
(2x^3 + 1)(2y^3 + 1)(2z^3 + 1) \ge (3\sqrt[3]{x^3 y^3 z^3})^3 = 27x^2 y^2 z^2
\]
Given the equations, we have:
\[
(2x^3 + 1)(2y^3 + 1)(2z^3 + 1) = (3zx)(3xy)(3yz) = 27x^2 y^2 z^2
\]
Equality in AM-GM holds if and only if \(2x^3 + 1 = 2y^3 + 1 = 2z^3 + 1\). This implies \(x = y = z\). Substituting \(x = y = z\) into any of the original equations, we get:
\[
2x^3 + 1 = 3x^2
\]
Solving this equation:
\[
2x^3 + 1 = 3x^2 \implies 2x^3 - 3x^2 + 1 = 0 \implies (x-1)^2(2x+1) = 0
\]
Thus, \(x = 1\) or \(x = -\frac{1}{2}\). Since we assumed \(x, y, z \ge 0\), the only solution in this case is \(x = y = z = 1\).

### Case 2: Exactly one of \(x, y, z\) is greater than 0
Without loss of generality, assume \(x > 0\) and \(y, z \le 0\). From the first equation:
\[
2x^3 + 1 > 0 \implies 3zx > 0 \implies z > 0
\]
This is a contradiction since we assumed \(z \le 0\). Therefore, there are no solutions in this case.

### Case 3: Exactly two of \(x, y, z\) are greater than 0
Without loss of generality, assume \(x, y > 0\) and \(z \le 0\). From the first equation:
\[
2x^3 + 1 > 0 \implies 3zx > 0 \implies z > 0
\]
This is a contradiction since we assumed \(z \le 0\). Therefore, there are no solutions in this case either.

### Case 4: \(x, y, z < 0\)
Assume \(x = \max\{x, y, z\}\). From the first equation:
\[
2x^3 + 1 \ge 2z^3 + 1 \implies 3zx \ge 3yz \implies x \le y
\]
This implies \(x = y\). Now, we have \(x = y \ge z\). From the second equation:
\[
2y^3 + 1 \ge 2z^3 + 1 \implies 3xy \ge 3yz \implies x \le z
\]
This implies \(x = z\). Therefore, \(x = y = z\). Substituting \(x = y = z\) into any of the original equations, we get:
\[
2x^3 + 1 = 3x^2
\]
Solving this equation:
\[
2x^3 + 1 = 3x^2 \implies 2x^3 - 3x^2 + 1 = 0 \implies (x-1)^2(2x+1) = 0
\]
Thus, \(x = 1\) or \(x = -\frac{1}{2}\). Since we assumed \(x, y, z < 0\), the only solution in this case is \(x = y = z = -\frac{1}{2}\).

The final answer is \( \boxed{ (1, 1, 1) } \) and \(\left(-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}\right)\).