If $s=x^2+y^2+z^2$ and $p=xyz$, we have by QM-GM $p\leq |p| \leq\left(\frac{s}{3}\right)^{3/2}$. So $1=s+2p\leq s+2\left(\frac{s}{3}\right)^{3/2}$ Because RHS is an increasing function and $s=\frac34$ satisfies the inequality, it follows $s\geq \frac34$ Therefore $8xyz=8p=4(1-s)\geq 4\left(1-\frac34\right)=1$ Equality hold iff ,$p=|p|$ and $|x|=|y|=|z|$ by equality case of QM-GM. So there's an even number of negatives, but wlog we can take them all positives, so we get the equation $3x^2+2x^3=1$ which has as only positive solution $x=\frac12$ because LHS is increasing. Therefore the only possibilities are $\left(\frac12,\frac12,\frac12\right),\left(\frac12,\frac12,-\frac12\right),\left(\frac12,-\frac12,\frac12\right),\left(-\frac12,\frac12,\frac12\right)$