(a) Differentiate \(y = 3x + 1 - 4(3x + 1)^{\frac{1}{2}}\).

\(\frac{dy}{dx} = 3 - 4 \times \frac{1}{2} \times (3x+1)^{-\frac{1}{2}} \times 3\)

\(\frac{dy}{dx} = 3 - 6(3x+1)^{-\frac{1}{2}}\)

For \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx}\):

\(\frac{d^2y}{dx^2} = -6 \times -\frac{1}{2} \times (3x+1)^{-\frac{3}{2}} \times 3\)

\(\frac{d^2y}{dx^2} = 9(3x+1)^{-\frac{3}{2}}\)

(b) Set \(\frac{dy}{dx} = 0\):

\(3 - 6(3x+1)^{-\frac{1}{2}} = 0\)

\((3x+1)^{-\frac{1}{2}} = \frac{1}{2}\)

\(3x+1 = 4\)

\(x = 1\)

Substitute \(x = 1\) into \(y = 3x + 1 - 4(3x + 1)^{\frac{1}{2}}\):

\(y = 3(1) + 1 - 4(4)^{\frac{1}{2}}\)

\(y = 3 + 1 - 4 \times 2\)

\(y = -4\)

Coordinates: (1, -4)

Check nature using \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = 9(3 \times 1 + 1)^{-\frac{3}{2}} = \frac{9}{8} > 0\)

Therefore, the stationary point is a minimum.