To find the stationary points, we first differentiate the equation \(y = \frac{k^2}{x+2} + x\) with respect to \(x\):

\(\frac{dy}{dx} = -\frac{k^2}{(x+2)^2} + 1\).

Set \(\frac{dy}{dx} = 0\) to find the stationary points:

\(-\frac{k^2}{(x+2)^2} + 1 = 0\).

Solving for \(x\), we get:

\((x+2)^2 = k^2\).

\(x+2 = \pm k\).

Thus, \(x = -2 + k\) or \(x = -2 - k\).

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

\(\frac{d^2y}{dx^2} = 2k^2(x+2)^{-3}\).

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

\(\frac{d^2y}{dx^2} = \frac{2}{k}\), which is positive, indicating a minimum.

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

\(\frac{d^2y}{dx^2} = \frac{-2}{k}\), which is negative, indicating a maximum.