Let's define $f(t) = \frac{t}{\sqrt{t^4 + t^2 + 1}}$, that is, $f(t) = -f(-t)$, $f(\frac{1}{t}) = f(t)$, furthermore, $|f(t)| \le \frac{1}{\sqrt{3}}$. Then, if we change $(x, y, z)$ with $(\frac{1}{x}, \frac{1}{y}, \frac{1}{z})$, nothing has been changed. Moreover, one can find that, $z = \frac{x + y - xy}{x + y - 1}$, so if $x, y > 0$ and $z < 0$ then
$$\frac{x}{\sqrt{x^4 + x^2 + 1}} + \frac{y}{\sqrt{y^4 + y^2 + 1}} + \frac{z}{\sqrt{z^4 + z^2 + 1}} \ge \frac{-1}{\sqrt{3}}$$
Moreover, if $x, y, z > 0$, we are done. Assume now, $x \le y \le 0 \le z$, Without loss of generality, we can assume $xy \ge 1$, otherwise, replace $(x, y, z)$ with $(\frac{1}{x}, \frac{1}{y}, \frac{1}{z})$. If $z = 0$, then $0 > x + y = xy > 0$, a contradiction, hence $z > 0$. Now, we can write
$$z = \frac{x + y - xy}{x + y - 1} = \frac{|x| + |y| + xy}{|x| + |y| + 1} \ge 1$$
Hence,
$$z - |x| = z + x = \frac{x^2 + y}{x + y - 1} = \frac{|y| - x^2}{|x + y| + 1}$$
Then, $1 \le xy \le x^2$ implies that $|x| \ge 1$. Moreover, $x \le y \le 0$ ensures that $|y| \le |x| \le x^2$, hence, $z - |x| \le 0$. Thus, $1 \le z \le |x|$. Finally, it is clear that
$$f(t) = \frac{1}{\sqrt{t^2 + t^{-2} + 1}}$$
The function, $t^2 + t^{-2}$ is increasing for $|t| \ge 1$, hence $f(t)$ is decreasing on $|t| \ge 1$, therefore $f(|x|) \le f(z)$. That is
$$-f(x) = f(-x) = f(|x|) \le f(z)$$
Hence $f(x) + f(z) \ge 0$ and $f(y) \ge -\frac{1}{\sqrt{3}}$, thus
$$f(x) + f(y) + f(z) \ge \frac{-1}{\sqrt{3}}.$$