To find the derivative \(\frac{dy}{dx}\) of \(y = \frac{e^{-2x}}{1-x^2}\), we use the quotient rule:

\(\frac{dy}{dx} = \frac{(1-x^2)(-2e^{-2x}) - e^{-2x}(-2x)}{(1-x^2)^2}\)

Simplifying, we get:

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

To find the stationary point, set \(\frac{dy}{dx} = 0\):

\(-2e^{-2x}(1-x^2) + 2xe^{-2x} = 0\)

Factor out \(e^{-2x}\):

\(e^{-2x}(-2(1-x^2) + 2x) = 0\)

Since \(e^{-2x} \neq 0\), solve:

\(-2 + 2x^2 + 2x = 0\)

Rearrange to form a quadratic equation:

\(2x^2 + 2x - 2 = 0\)

Divide by 2:

\(x^2 + x - 1 = 0\)

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = 1, c = -1\):

\(x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}\)

\(x = \frac{-1 \pm \sqrt{5}}{2}\)

Calculate the positive root within the interval \(-1 < x < 1\):

\(x = \frac{-1 + \sqrt{5}}{2} \approx 0.618\)