(i) To find the stationary point, we first find the derivative of \(y = \frac{\ln x}{x + 1}\) using the quotient rule:

\(\frac{d}{dx} \left( \frac{\ln x}{x + 1} \right) = \frac{(x + 1) \cdot \frac{1}{x} - \ln x \cdot 1}{(x + 1)^2} = \frac{1 - \ln x}{x(x + 1)^2}\).

Setting the derivative to zero gives:

\(1 - \ln x = 0\) or \(\ln x = 1\).

Solving \(\ln x = 1\) gives \(x = e\), but we need to verify the given equation:

Rearrange \(x = \frac{x + 1}{\ln x}\) to \(x \ln x = x + 1\), which simplifies to \(\ln x = 1 + \frac{1}{x}\).

Check the interval \(x = 3\) and \(x = 4\):

For \(x = 3\), \(\ln 3 \approx 1.0986\) and \(\frac{3 + 1}{3} \approx 1.3333\).

For \(x = 4\), \(\ln 4 \approx 1.3863\) and \(\frac{4 + 1}{4} = 1.25\).

Since \(\ln x\) changes from greater than \(\frac{x + 1}{x}\) to less than \(\frac{x + 1}{x}\) between 3 and 4, there is a root in this interval.

(ii) Using the iterative formula \(x_{n+1} = \frac{x_n + 1}{\ln x_n}\):

Start with \(x_1 = 3.5\):

\(x_2 = \frac{3.5 + 1}{\ln 3.5} \approx 3.5718\)

\(x_3 = \frac{3.5718 + 1}{\ln 3.5718} \approx 3.5911\)

\(x_4 = \frac{3.5911 + 1}{\ln 3.5911} \approx 3.5900\)

\(x_5 = \frac{3.5900 + 1}{\ln 3.5900} \approx 3.5900\)

The \(x\)-coordinate correct to 2 decimal places is 3.59.