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.