System

By observation, we see that neither $(0;0)$ , $(0;a)$ or $(b;0)$ are solutions to this problem ( $a$ and $b$ are reals) so now we set $y=kx \ (k\in R)$ , the system'll become $\begin{cases} 2(x^3-3x)=k^3x^3+3kx \\ 2(x^2-1)=5k^2x^2+1\end{cases}$ $\Leftrightarrow\begin{cases} x^2=\frac{3(2+k)}{2-k^3} \\ x^2=\frac{3}{2-5k^2}\end{cases}$ $\Leftrightarrow\begin{cases} \frac{2+k}{2-k^3}=\frac{1}{2-5k^2} \\ x^2=\frac{3}{2-5k^2}\end{cases}$ $\Leftrightarrow\begin{cases} 2k^3+5k^2-k-1=0 \\ x^2=\frac{3}{2-5k^2}\end{cases}$ From the first equation we get $k=\big\{\frac{1}{2};\frac{-3+\sqrt{5}}{2};\frac{-3-\sqrt{5}}{2}\big\}$ . Therefore

* $k=\frac{1}{2}$ we have two solutions $(x;y)=(2;1),(-2;-1)$

* $k=\frac{-3+\sqrt{5}}{2}$ we have two solutions $(x;y)=\bigg(\sqrt{\frac{3}{2-5k^2}};k\sqrt{\frac{3}{2-5k^2}}\bigg),\bigg(-\sqrt{\frac{3}{2-5k^2}};-k\sqrt{\frac{3}{2-5k^2}}\bigg)$

* $k=\frac{-3-\sqrt{5}}{2}$ we have no solution

So $\displaystyle \prod_{m=1}^n x_m y_m =2.1.(-2).(-1).k\frac{3}{2-5k^2}.\bigg(-\sqrt{\frac{3}{2-5k^2}}\bigg).\bigg(-k\sqrt{\frac{3}{2-5k^2}}\bigg)\approx 3.25 \ \big(k=\frac{-3+\sqrt{5}}{2}\big)$