Solve for $x, y \in \mathbb R$: $(x^2 - 1)^2 + 3 = \dfrac{6x^5y}{x^2 + 2}, 3y - x = \sqrt{\dfrac{4x - 3x^2y - 9xy^2}{x + 3y}}$ 
Solve the following system of equations $$\large \left\{ \begin{align} (x^2 &- 1)^2 + 3 = \frac{6x^5y}{x^2 + 2}\\ 3y - x &= \sqrt{\frac{4x - 3x^2y - 9xy^2}{x + 3y}}\end{align} \right.$$ such that $(x, y)$ is a root but $(-x, -y)$ is not.

I have provided a solution below. It looks disordered and there might be better solutions out there, I don't know.
 A: $(x \le 3y \ne -x)$
Let $a = \dfrac{2}{x^2} > 0$ and $b = \dfrac{3y}{x}$. The system of equation becomes
$$\left\{ \begin{align} \left(\frac{2}{a} - 1\right)^2 + 3 = \frac{2b \cdot \left(\dfrac{2}{a}\right)^3}{\dfrac{2}{a} + 2}\\ b - 1 = \sqrt{\frac{2a - b - b^2}{1 + b}} \end{align} \right.$$
$$\iff \left\{ \begin{align} (2 - a)^2 + 3a^2 = \frac{8b}{a + 1}\\ (b - 1)^2 + b = \frac{2a}{b + 1} \end{align} \right.$$
$$\iff \left\{ \begin{align} a^2 - a + 1 = \frac{2b}{a + 1}\\ b^2 - b + 1 = \frac{2a}{b + 1} \end{align} \right.$$
$$\iff \left\{ \begin{align} a^3 + 1 = 2b\\ b^3 + 1 = 2a \end{align} \right.$$
$\iff a^3 - 2b = b^3 - 2a \iff (a - b)(a^2 + ab + b^2 + 2) = 0$
However $a^2 + ab + b^2 + 2 \ge 2 > 0, \forall a, b \in \mathbb R$
$\implies a^3 - 2a + 1 = b^3 - 2b + 1 = 0$ and $a = b$
$\iff (a - 1)(a^2 + a - 1) = (b - 1)(b^2 + b - 1) = 0$ and $a = b$
$\iff a = b = 1$ or $a = b = \dfrac{\sqrt 5 - 1}{2} (a > 0)$
$\implies \dfrac{2}{x^2} = \dfrac{3x}{y} = 1$ or $\dfrac{2}{x^2} = \dfrac{3x}{y} = \dfrac{\sqrt 5 - 1}{2}$
$\implies (x, y) = \left(\pm \sqrt 2, \pm \dfrac{\sqrt 2}{3}\right)$ or $(x, y) = \left(\pm \sqrt{\sqrt 5 + 1}, \pm \dfrac{\sqrt{\sqrt 5 + 1}\left(\sqrt 5 - 1\right)}{6}\right)$
Nevertheless $x \le 3y$, $\implies \left(- \sqrt{\sqrt 5 + 1}, - \dfrac{\sqrt{\sqrt 5 - 1}}{3}\right)$ is a solution but $\left(\sqrt{\sqrt 5 + 1}, \dfrac{\sqrt{\sqrt 5 - 1}}{3}\right)$ is not.
