Using the quotient rule:

\[ f'(x) = \frac{x \cdot \frac{1}{x} – \ln x \cdot 1}{x^2} = \frac{1 – \ln x}{x^2} \]

\(\boxed{f'(x) = \frac{1 – \ln x}{x^2}}\)

Set \(f'(x) = 0\):

\[ \frac{1 – \ln x}{x^2} = 0 \Rightarrow 1 – \ln x = 0 \Rightarrow \ln x = 1 \Rightarrow x = e \]

Find \(y\)-coordinate:

\[ f(e) = \frac{\ln e}{e} = \frac{1}{e} \]

Maximum at point B:

\(\boxed{\left(e, \frac{1}{e}\right)}\)

Find second derivative:

\[ f”(x) = \frac{x^2(-\frac{1}{x}) – (1 – \ln x)(2x)}{x^4} = \frac{-x – 2x + 2x\ln x}{x^4} = \frac{2\ln x – 3}{x^3} \]

Set \(f”(x) = 0\):

\[ 2\ln x – 3 = 0 \Rightarrow \ln x = \frac{3}{2} \Rightarrow x = e^{3/2} \]

Find \(y\)-coordinate:

\[ f(e^{3/2}) = \frac{3/2}{e^{3/2}} = \frac{3}{2}e^{-3/2} \]

Point of inflexion C:

\(\boxed{\left(e^{3/2}, \frac{3}{2}e^{-3/2}\right)}\)

Find point A where \(f(x) = 0\):

\[ \frac{\ln x}{x} = 0 \Rightarrow \ln x = 0 \Rightarrow x = 1 \]

Point A is \((1, 0)\).

Slope at A:

\[ f'(1) = \frac{1 – \ln 1}{1} = 1 \]

Tangent equation:

\[ y – 0 = 1(x – 1) \Rightarrow y = x – 1 \]

\(\boxed{y = x – 1}\)

Area between \(x = 1\) and \(x = e\):

\[ \text{Area} = \int_1^e \left[(x – 1) – \frac{\ln x}{x}\right] dx \]

Calculate integrals separately:

\[ \int (x – 1) dx = \frac{x^2}{2} – x \]

\[ \int \frac{\ln x}{x} dx = \frac{(\ln x)^2}{2} \]

Evaluate from 1 to e:

\[ \left[\frac{x^2}{2} – x – \frac{(\ln x)^2}{2}\right]_1^e = \left(\frac{e^2}{2} – e – \frac{1}{2}\right) – \left(\frac{1}{2} – 1\right) = \frac{e^2}{2} – e \]

\(\boxed{\frac{e^2}{2} – e}\)