To find the stationary points, we need to find where the derivative \(\frac{dy}{dx}\) is zero.

Using the quotient rule, the derivative is:

\(\frac{dy}{dx} = \frac{6x(1-x^2)e^{3x^2-1} + 2xe^{3x^2-1}}{(1-x^2)^2}\).

Simplifying the numerator, we have:

\(6x(1-x^2) + 2x = 6x - 6x^3 + 2x = 8x - 6x^3\).

Setting the numerator equal to zero gives:

\(8x - 6x^3 = 0\).

Factoring out \(2x\), we get:

\(2x(4 - 3x^2) = 0\).

This gives solutions \(x = 0\) or \(4 - 3x^2 = 0\).

Solving \(4 - 3x^2 = 0\) gives \(x^2 = \frac{4}{3}\), so \(x = \pm \frac{2\sqrt{3}}{3}\).

Substituting these \(x\)-values back into the original equation to find \(y\):

For \(x = 0\), \(y = e^{-1}\).

For \(x = \pm \frac{2\sqrt{3}}{3}\), \(y = -3e^3\).

Thus, the stationary points are \((0, e^{-1})\), \(\left( \frac{2\sqrt{3}}{3}, -3e^3 \right)\), and \(\left( -\frac{2\sqrt{3}}{3}, -3e^3 \right)\).