The quotient rule gives \(y^{\prime}=\dfrac{6x(1-x^2)e^{3x^2-1}+2xe^{3x^2-1}}{(1-x^2)^2}=\dfrac{2xe^{3x^2-1}(4-3x^2)}{(1-x^2)^2}\).

The exponential is nonzero, and \(x=\pm1\) are outside the domain. Therefore \(x=0,\ \pm2/\sqrt3\). The corresponding ordinates are \(e^{-1},-3e^3,-3e^3\). The points are \(\boxed{(0,e^{-1}),\ (2\sqrt3/3,-3e^3),\ (-2\sqrt3/3,-3e^3)}\).