Find all real solutions to: \begin{eqnarray*} 3(x^2 + y^2 + z^2) &=& 1 \\ x^2y^2 + y^2z^2 + z^2x^2 &=& xyz(x + y + z)^3. \end{eqnarray*}
1. Given the system of equations:
   \[
   3(x^2 + y^2 + z^2) = 1
   \]
   \[
   x^2y^2 + y^2z^2 + z^2x^2 = xyz(x + y + z)^3
   \]

2. First, we use the Cauchy-Schwarz inequality in the form:
   \[
   (x^2 + y^2 + z^2)(1 + 1 + 1) \geq (x + y + z)^2
   \]
   Substituting \(3(x^2 + y^2 + z^2) = 1\), we get:
   \[
   1 \geq (x + y + z)^2
   \]
   Therefore:
   \[
   |x + y + z| \leq 1
   \]

3. Next, consider the second equation:
   \[
   x^2y^2 + y^2z^2 + z^2x^2 = xyz(x + y + z)^3
   \]
   We can rewrite it as:
   \[
   x^2y^2 + y^2z^2 + z^2x^2 \geq xyz(x + y + z)
   \]
   This follows from the AM-GM inequality:
   \[
   x^2y^2 + y^2z^2 + z^2x^2 \geq 3\sqrt[3]{(x^2y^2)(y^2z^2)(z^2x^2)} = 3x^2y^2z^2
   \]
   Since \(x, y, z\) are real numbers, we have:
   \[
   x^2y^2 + y^2z^2 + z^2x^2 \geq 3x^2y^2z^2
   \]

4. Combining the inequalities, we get:
   \[
   xyz(x + y + z)^3 \leq x^2y^2 + y^2z^2 + z^2x^2 \leq 3x^2y^2z^2
   \]
   This implies:
   \[
   xyz(x + y + z)^3 \leq 3x^2y^2z^2
   \]

5. For equality to hold in the AM-GM inequality, we must have:
   \[
   x^2 = y^2 = z^2
   \]
   Since \(x, y, z\) are real numbers, we can assume \(x = y = z\).

6. Substituting \(x = y = z\) into the first equation:
   \[
   3(x^2 + x^2 + x^2) = 1 \implies 9x^2 = 1 \implies x^2 = \frac{1}{9} \implies x = \pm \frac{1}{3}
   \]

7. Substituting \(x = y = z = \frac{1}{3}\) into the second equation:
   \[
   \left(\frac{1}{3}\right)^2 \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 \left(\frac{1}{3}\right)^2 = \left(\frac{1}{3}\right)\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)\left(\frac{1}{3} + \frac{1}{3} + \frac{1}{3}\right)^3
   \]
   Simplifying:
   \[
   3 \left(\frac{1}{9}\right) = \left(\frac{1}{27}\right) \left(1\right)^3 \implies \frac{1}{3} = \frac{1}{3}
   \]
   This holds true.

8. Similarly, substituting \(x = y = z = -\frac{1}{3}\) into the second equation:
   \[
   \left(-\frac{1}{3}\right)^2 \left(-\frac{1}{3}\right)^2 + \left(-\frac{1}{3}\right)^2 \left(-\frac{1}{3}\right)^2 + \left(-\frac{1}{3}\right)^2 \left(-\frac{1}{3}\right)^2 = \left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right)\left(-\frac{1}{3} - \frac{1}{3} - \frac{1}{3}\right)^3
   \]
   Simplifying:
   \[
   3 \left(\frac{1}{9}\right) = \left(-\frac{1}{27}\right) \left(-1\right)^3 \implies \frac{1}{3} = \frac{1}{3}
   \]
   This also holds true.

Therefore, the solutions are \(x = y = z = \frac{1}{3}\) and \(x = y = z = -\frac{1}{3}\).

The final answer is \(\boxed{x = y = z = \pm \frac{1}{3}}\).