(a) To find the stationary point, we first find the derivative of \(y = xe^{1-2x}\) using the product rule. Let \(u = x\) and \(v = e^{1-2x}\). Then \(\frac{du}{dx} = 1\) and \(\frac{dv}{dx} = -2e^{1-2x}\).

Using the product rule, \(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = x(-2e^{1-2x}) + e^{1-2x}\).

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

Set \(\frac{dy}{dx} = 0\) to find the stationary point: \(e^{1-2x}(1 - 2x) = 0\).

Since \(e^{1-2x} \neq 0\), we have \(1 - 2x = 0\), giving \(x = \frac{1}{2}\).

Substitute \(x = \frac{1}{2}\) back into the original equation to find \(y\):

\(y = \frac{1}{2}e^{1-2(\frac{1}{2})} = \frac{1}{2}e^{0} = \frac{1}{2}\).

Thus, the coordinates of the stationary point are \(\left( \frac{1}{2}, \frac{1}{2} \right)\).

(b) To determine the nature of the stationary point, we find the second derivative \(\frac{d^2y}{dx^2}\).

Differentiate \(\frac{dy}{dx} = e^{1-2x}(1 - 2x)\) again using the product rule:

\(\frac{d^2y}{dx^2} = \frac{d}{dx}(e^{1-2x}(1 - 2x)) = -2e^{1-2x}(1 - 2x) - 2e^{1-2x}\).

At \(x = \frac{1}{2}\), \(\frac{d^2y}{dx^2} = -2e^{1-2(\frac{1}{2})}(1 - 2(\frac{1}{2})) - 2e^{1-2(\frac{1}{2})} = -2e^{0}(0) - 2e^{0} = -2\).

Since \(\frac{d^2y}{dx^2} < 0\), the stationary point is a maximum.