We will show that for real numbers $x, y,$ and $z$ that satisfy the first condition, the following inequality holds:

$$x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2} \geq x y z (x + y + z)^{3}.$$

The solutions we are looking for are thus the equality cases of this inequality.
Let $a, b,$ and $c$ be real numbers. It holds that $(a - b)^{2} \geq 0$, so $\frac{a^{2} + b^{2}}{2} \geq a b$ with equality if and only if $a = b$. We do this analogously for $b$ and $c$ and for $c$ and $a$, so that adding them up we get

$$a^{2} + b^{2} + c^{2} \geq a b + b c + c a$$

with equality if and only if $a = b = c$.
If we apply this to $(x y, y z, z x)$, we find that

$$x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2} \geq x y^{2} z + y z^{2} x + z x^{2} y = x y z (x + y + z)$$

with equality if and only if $x y = y z = z x$. By applying it to $(x, y, z)$, we get $x^{2} + y^{2} + z^{2} \geq x y + y z + z x$, so according to the first condition,

$$1 = 3 (x^{2} + y^{2} + z^{2}) = (x^{2} + y^{2} + z^{2}) + 2 (x^{2} + y^{2} + z^{2}) \geq x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x = (x + y + z)^{2},$$

with equality if and only if $x = y = z$. We now want to combine these two inequalities,

$$x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2} \geq x y z (x + y + z)$$

and

$$1 \geq (x + y + z)^{2}$$

with each other. For this, we first multiply (3) by the non-negative factor $x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2}$; then we get

$$(x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2}) \cdot 1 \geq (x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2}) \cdot (x + y + z)^{2}$$

with equality if and only if $x = y = z$ or $x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2} = 0$. Further, multiplying (2) by the non-negative factor $(x + y + z)^{2}$, we get

$$(x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2}) \cdot (x + y + z)^{2} \geq x y z (x + y + z) \cdot (x + y + z)^{2}$$

with equality if and only if $x y = y z = z x$ or $x + y + z = 0$. Altogether, we find $(x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2}) \cdot 1 \geq (x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2}) \cdot (x + y + z)^{2} \geq x y z (x + y + z) \cdot (x + y + z)^{2}$,
or in other words (1), with equality if and only if

$$\left(x = y = z \text{ or } x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2} = 0\right) \text{ and } (x y = y z = z x \text{ or } x + y + z = 0).$$

Now the second equation says that equality must hold here, so we are in the equality case just described. Additionally, the first condition still states that $3 (x^{2} + y^{2} + z^{2}) = 1$:

$$\begin{aligned} & 3 (x^{2} + y^{2} + z^{2}) = 1 \\ & x = y = z \text{ or } \\ & x y = y z = z x \text{ or } \\ & x + y + y^{2} z^{2} + z^{2} x^{2} = 0 \\ & x y = 0 \end{aligned}$$

We distinguish two cases. First, assume $x = y = z$. Then $x y = y z = z x$ is automatically satisfied. Using the first condition, we find that $1 = 3 (x^{2} + y^{2} + z^{2}) = 9 x^{2} = (3 x)^{2}$, so $x = y = z = \pm \frac{1}{3}$.
Now assume that $x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2} = 0$; then each of the terms must be zero, so $x y = y z = z x = 0$, so $x y = y z = z x$ is also satisfied. From $x y = 0$ it follows that (without loss of generality) $x = 0$. From $y z = 0$ it also follows that (without loss of generality) $y = 0$. Thus we find the solutions $\left(0, 0, \pm \frac{1}{3} \sqrt{3}\right)$ and also $\left(0, \pm \frac{1}{3} \sqrt{3}, 0\right)$ and $\left(\pm \frac{1}{3} \sqrt{3}, 0, 0\right)$.

Remark. If you have proven (2) and (3), you can also find (1) by multiplying (3) by $x y z (x + y + z)$. For this, you need to show that $x y z (x + y + z)$ is non-negative. We know from the second condition in the problem that $x y z (x + y + z)^{3}$ is equal to a sum of squares and is therefore non-negative. If $x + y + z \neq 0$, it follows that $x y z (x + y + z) \geq 0$. If $x + y + z = 0$, then $x y z (x + y + z) = 0$. So in both cases, $x y z (x + y + z) \geq 0$.
The equality case for this method is:

$$\begin{aligned} & 3 (x^{2} + y^{2} + z^{2}) = 1 \\ & \quad x y = y z = z x \\ & x = y = z \quad \text{or} \quad x y z (x + y + z) = 0. \end{aligned}$$