(i) Differentiate \(y = \frac{\ln x}{3 + x}\) using the quotient rule:

\(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\)

Let \(u = \ln x\) and \(v = 3 + x\). Then \(\frac{du}{dx} = \frac{1}{x}\) and \(\frac{dv}{dx} = 1\).

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

Set \(\frac{dy}{dx} = 0\) for stationary points:

\(3 + x - x \ln x = 0\)

\(x \ln x = 3 + x\)

\(\ln x = 1 + \frac{3}{x}\)

(ii) Sketch the graphs of \(y = \ln x\) and \(y = 1 + \frac{3}{x}\). The intersection point represents the root. Since \(\ln x\) is a monotonically increasing function and \(1 + \frac{3}{x}\) is decreasing for \(x > 0\), they intersect at only one point.

(iii) Use the iterative formula \(x_{n+1} = \frac{3 + x_n}{\ln x_n}\):

Start with an initial guess, say \(x_1 = 4.5\).

Calculate successive iterations:

\(x_2 = \frac{3 + 4.5}{\ln 4.5} \approx 4.9686\)

\(x_3 = \frac{3 + 4.9686}{\ln 4.9686} \approx 4.9701\)

\(x_4 = \frac{3 + 4.9701}{\ln 4.9701} \approx 4.9700\)

Continue until the value stabilizes to 4 decimal places. The final answer is 4.97.