If the tangent to the curve y=x/(x^2-3), x ∈ R,(x ≠ ± √ 3 ), at a point (α, β) ≠(0,0) on it is parallel to the line 2 x+6 y-11=0, then:

If the tangent to the curve $\displaystyle y=\frac{x}{x^2-3}$, $\displaystyle x \in \mathbf{R},(x \neq \pm \sqrt{3})$, at a point $\displaystyle (\alpha, \beta) \neq(0,0)$ on it is parallel to the line $\displaystyle 2 x+6 y-11=0$, then :