Note the following observations: (a) if $(x, y)$ is a solution then $(\omega x, \omega^{2} y)$ is also a solution if $\omega^{3}=1$ and $\omega \neq 1$. (b) we have some solutions $(x, x)$ where $x$ is a solution of $x^{4}-2x^{2}-x=0$. These are really the only necessary observations and the first does not need to be noticed immediately. Indeed, we can try to solve this directly as follows: first, from the first equation, we get $y^{2}=\frac{1}{2}(x^{4}-x)$, so inserting this into the second equation gives $$\begin{aligned} \frac{1}{4}(x^{4}-x)^{2}-2x^{2} & =y \\ \left((x^{4}-x)^{2}-8x^{2}\right)^{2}-8x^{4}+8x & =0 \\ x^{16}+\cdots+41x^{4}+8x & =0 \end{aligned}$$ By the second observation, we have that $x(x^{3}-2x-1)$ should be a factor of $P$. The first observation gives that $(x^{3}-2\omega x-1)(x^{3}-2\omega^{2} x-1)$ should therefore also be a factor. Now $(x^{3}-2\omega x-1)(x^{3}-2\omega^{2} x-1)=x^{6}+2x^{4}-2x^{3}+4x^{2}-2x+1$ since $\omega$ and $\omega^{2}$ are roots of $x^{2}+x+1$. So now we see that the last two terms of the product of all of these is $-5x^{4}-x$. Hence the last two terms of the polynomial we get after dividing out should be $-x^{3}-8$, and given what we know about the degree and the fact that everything is monic, the quotient must be exactly $x^{6}-x^{3}-8$ which has roots being the cube roots of the roots to $x^{2}-x-8$, which are $\sqrt[3]{\frac{1 \pm \sqrt{33}}{2}}$. Now $x^{3}-2x-1$ is further factorable as $(x-1)(x^{2}-x-1)$ with roots $1, \frac{1 \pm \sqrt{5}}{2}$ so it is not difficult to compare the real parts of all roots of $P$, especially since 5 are real and non-zero, and we have that $\operatorname{Re}(\omega x)=-\frac{1}{2} x$ if $x \in \mathbb{R}$. We conclude that the smallest is $\sqrt[3]{\frac{1-\sqrt{33}}{2}}$.