Solve in the set of real numbers, the system:
$$x(3y^2+1)=y(y^2+3)$$
$$y(3z^2+1)=z(z^2+3)$$
$$z(3x^2+1)=x(x^2+3)$$
To solve the system of equations in the set of real numbers:

\[
\begin{cases}
x(3y^2 + 1) = y(y^2 + 3) \\
y(3z^2 + 1) = z(z^2 + 3) \\
z(3x^2 + 1) = x(x^2 + 3)
\end{cases}
\]

we will analyze each equation and look for possible solutions.

1. **Analyzing the first equation:**

\[
x(3y^2 + 1) = y(y^2 + 3)
\]

Rewriting it, we get:

\[
x(3y^2 + 1) - y(y^2 + 3) = 0
\]

This can be factored as:

\[
x(3y^2 + 1) - y(y^2 + 3) = 0 \implies x(3y^2 + 1) - y^3 - 3y = 0
\]

2. **Analyzing the second equation:**

\[
y(3z^2 + 1) = z(z^2 + 3)
\]

Rewriting it, we get:

\[
y(3z^2 + 1) - z(z^2 + 3) = 0
\]

This can be factored as:

\[
y(3z^2 + 1) - z^3 - 3z = 0 \implies y(3z^2 + 1) - z^3 - 3z = 0
\]

3. **Analyzing the third equation:**

\[
z(3x^2 + 1) = x(x^2 + 3)
\]

Rewriting it, we get:

\[
z(3x^2 + 1) - x(x^2 + 3) = 0
\]

This can be factored as:

\[
z(3x^2 + 1) - x^3 - 3x = 0 \implies z(3x^2 + 1) - x^3 - 3x = 0
\]

4. **Finding common solutions:**

We observe that if \(x = y = z\), then each equation becomes:

\[
x(3x^2 + 1) = x(x^2 + 3)
\]

Simplifying, we get:

\[
3x^3 + x = x^3 + 3x \implies 2x^3 - 2x = 0 \implies 2x(x^2 - 1) = 0 \implies x(x - 1)(x + 1) = 0
\]

Thus, the possible solutions for \(x\) are:

\[
x = 0, \quad x = 1, \quad x = -1
\]

Since \(x = y = z\), the corresponding solutions for \((x, y, z)\) are:

\[
(0, 0, 0), \quad (1, 1, 1), \quad (-1, -1, -1)
\]

5. **Verifying the solutions:**

- For \((0, 0, 0)\):

\[
0(3 \cdot 0^2 + 1) = 0 \cdot 3 = 0 \quad \text{(True)}
\]

- For \((1, 1, 1)\):

\[
1(3 \cdot 1^2 + 1) = 1 \cdot 4 = 4 \quad \text{and} \quad 1(1^2 + 3) = 1 \cdot 4 = 4 \quad \text{(True)}
\]

- For \((-1, -1, -1)\):

\[
-1(3 \cdot (-1)^2 + 1) = -1 \cdot 4 = -4 \quad \text{and} \quad -1((-1)^2 + 3) = -1 \cdot 4 = -4 \quad \text{(True)}
\]

Thus, the real solutions are:

\[
\boxed{(1, 1, 1), (-1, -1, -1), (0, 0, 0)}
\]