Find all triples $(x,y, z)$ of real (but not necessarily positive) numbers satisfying 
$3(x^2 + y^2 + z^2) = 1$ , $x^2y^2 + y^2z^2 + z^2x^2 = xyz(x + y + z)^3$.
1. **Given Equations:**
   \[
   3(x^2 + y^2 + z^2) = 1
   \]
   \[
   x^2y^2 + y^2z^2 + z^2x^2 = xyz(x + y + z)^3
   \]

2. **Check the proposed solutions:**
   - \((x, y, z) = \left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)\)
   - \((x, y, z) = \left(-\frac{1}{3}, -\frac{1}{3}, -\frac{1}{3}\right)\)
   - \((x, y, z) = \left(\frac{1}{\sqrt{3}}, 0, 0\right)\) and its cyclic permutations.

3. **Verification of \(\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)\):**
   - Substitute into the first equation:
     \[
     3\left(\left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2\right) = 3\left(\frac{1}{9} + \frac{1}{9} + \frac{1}{9}\right) = 3 \cdot \frac{3}{9} = 1
     \]
     This satisfies the first equation.
   - Substitute into the second equation:
     \[
     \left(\frac{1}{3}\right)^2 \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 \left(\frac{1}{3}\right)^2 = \left(\frac{1}{3}\right)^3 \left(\frac{1}{3} + \frac{1}{3} + \frac{1}{3}\right)^3
     \]
     \[
     3 \cdot \left(\frac{1}{9}\right)^2 = \left(\frac{1}{27}\right) \left(1\right)^3
     \]
     \[
     3 \cdot \frac{1}{81} = \frac{1}{27}
     \]
     \[
     \frac{3}{81} = \frac{1}{27}
     \]
     This satisfies the second equation.

4. **Verification of \(\left(-\frac{1}{3}, -\frac{1}{3}, -\frac{1}{3}\right)\):**
   - The calculations are identical to the previous case since squaring the values will yield the same results.

5. **Verification of \(\left(\frac{1}{\sqrt{3}}, 0, 0\right)\):**
   - Substitute into the first equation:
     \[
     3\left(\left(\frac{1}{\sqrt{3}}\right)^2 + 0^2 + 0^2\right) = 3 \cdot \frac{1}{3} = 1
     \]
     This satisfies the first equation.
   - Substitute into the second equation:
     \[
     \left(\frac{1}{\sqrt{3}}\right)^2 \cdot 0^2 + 0^2 \cdot 0^2 + 0^2 \cdot \left(\frac{1}{\sqrt{3}}\right)^2 = \left(\frac{1}{\sqrt{3}}\right) \cdot 0 \cdot \left(\frac{1}{\sqrt{3}} + 0 + 0\right)^3
     \]
     \[
     0 = 0
     \]
     This satisfies the second equation.

6. **General Case Analysis:**
   - From the second equation, we have:
     \[
     x^2y^2 + y^2z^2 + z^2x^2 \geq xyz(x + y + z)
     \]
     This is equivalent to:
     \[
     (xy - yz)^2 + (yz - zx)^2 + (zx - xy)^2 \geq 0
     \]
     which is always true.

   - Therefore, \(xyz(x + y + z)^3 \geq xyz(x + y + z)\). If \(xyz(x + y + z)\) is positive, this implies:
     \[
     (x + y + z)^2 \geq 1
     \]
     Using the first equation:
     \[
     x^2 + y^2 + z^2 = \frac{1}{3}
     \]
     and the identity:
     \[
     (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)
     \]
     we get:
     \[
     \frac{1}{3} + 2(xy + yz + zx) \geq 1
     \]
     \[
     2(xy + yz + zx) \geq \frac{2}{3}
     \]
     \[
     xy + yz + zx \geq \frac{1}{3}
     \]
     Since \(x^2 + y^2 + z^2 \geq xy + yz + zx\), we have:
     \[
     xy + yz + zx \leq \frac{1}{3}
     \]
     Therefore:
     \[
     xy + yz + zx = \frac{1}{3}
     \]
     and equality holds, implying \(x = y = z\).

   - Solving \(3x^2 = \frac{1}{3}\):
     \[
     9x^2 = 1
     \]
     \[
     x = \pm \frac{1}{3}
     \]

   - If one of \(x, y, z\) is zero, say \(z = 0\), then:
     \[
     x^2y^2 = 0
     \]
     WLOG, let \(y = 0\), then:
     \[
     x = \pm \frac{1}{\sqrt{3}}
     \]

The final answer is \( \boxed{ \left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right), \left(-\frac{1}{3}, -\frac{1}{3}, -\frac{1}{3}\right), \left(\frac{1}{\sqrt{3}}, 0, 0\right) } \) and its cyclic permutations.