(a) To find \(\frac{dy}{dx}\), differentiate \(y = 2x + 1 + \frac{1}{2x+1}\).

The derivative of \(2x + 1\) is 2.

For \(\frac{1}{2x+1}\), use the chain rule: \(\frac{d}{dx} \left( (2x+1)^{-1} \right) = -1(2x+1)^{-2} \cdot 2 = -2(2x+1)^{-2}\).

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

To find \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx} = 2 - 2(2x+1)^{-2}\).

The derivative of \(-2(2x+1)^{-2}\) is \(4(2x+1)^{-3} \cdot 2 = 8(2x+1)^{-3}\).

Thus, \(\frac{d^2y}{dx^2} = 8(2x+1)^{-3}\).

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

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

\((2x+1)^{-2} = 1\) implies \(2x+1 = \pm 1\).

Solving \(2x+1 = 1\), we get \(x = 0\).

Substitute \(x = 0\) into \(y = 2x + 1 + \frac{1}{2x+1}\) to find \(y = 2(0) + 1 + \frac{1}{1} = 2\).

Thus, the stationary point is \((0, 2)\).

Since \(\frac{d^2y}{dx^2} = 8(2x+1)^{-3} > 0\) for \(x > -\frac{1}{2}\), the stationary point is a minimum.