To find the stationary point, we first differentiate the equation \(y = 2x + \frac{1}{(x-1)^2}\).

The derivative is \(\frac{dy}{dx} = 2 - 2(x-1)^{-3}\).

Substitute \(x = 2\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = 2 - 2(2-1)^{-3} = 2 - 2 = 0\).

This confirms a stationary point at \(x = 2\).

Next, we find the second derivative to determine the nature of the stationary point:

\(\frac{d^2y}{dx^2} = 6(x-1)^{-4}\).

Substitute \(x = 2\) into \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = 6(2-1)^{-4} = 6\).

Since \(\frac{d^2y}{dx^2} > 0\), the stationary point at \(x = 2\) is a minimum.