（1）由题意得\( f(x) = x(1 – \ln x) \)，定义域\( x \in (1, +\infty) \)

\( f'(x) = 1 – (\ln x + 1) = -\ln x \)

令\( f'(x) > 0 \) 即\( -\ln x > 0 \) ，得\( x \in (0,1) \)

\( \therefore f(x) \)在\( (0,1] \)上单调递增，在\( (1, +\infty) \)上单调递减。

（2）由\( b\ln a – a\ln b = a – b \)得

\( \frac{1}{a}\left(1 – \ln \frac{1}{a}\right) = \frac{1}{b}\left(1 – \ln \frac{1}{b}\right) \)， 即\( f\left(\frac{1}{a}\right) = f\left(\frac{1}{b}\right) \)

①证明\(\dfrac{1}{a} + \dfrac{1}{b} > 2\)（极值点偏移法）

设\(x_1 = \dfrac{1}{a}\)，\(x_2 = \dfrac{1}{b}\) 且 \(0 < x_1 < 1 < x_2 < e\)

即证\(x_1 + x_2 > 2\)，即证\(x_2 > 2 – x_1\)

\(\because x_2 \in (1, e)\) ，\(\therefore 2 – x_1 \in (1, 2)\)，\(\therefore\)只需证\(f(x_2) < f(2 - x_1)\)

\(\because f(x_1) = f(x_2)\)，\(\therefore\)只需证\(f(x_1) < f(2 - x_1)\)

设\(F(x) = f(x) – f(2 – x)\)

\(\therefore\)只需证\(F(x) = f(x) – f(2 – x) < 0\) 在定义域\((0, 1)\)上恒成立

\(F'(x) = f'(x) + f'(2 – x) = -\ln x – \ln(2 – x)\)

\(F”(x) = -\dfrac{1}{x} + \dfrac{1}{2 – x} = \dfrac{2x – 2}{x(2 – x)}\)

由\(x\)的取值范围可知，\(F”(x) < 0\) ，\(\therefore F'(x)\)在\((0, 1)\)单调递减，\(F'(1) = 0\)，\(F'(x) > F'(1) = 0\)

\(\therefore F'(x)\)在\((0, 1)\)单调递增且\(F(1) = 0\) ，\(\therefore F(x) < F(1) = 0\)

\(\therefore f(x) – f(2 – x)\) 在定义域\((0, 1)\)上恒成立

所以，\(\dfrac{1}{a} + \dfrac{1}{b} > 2\)成立

②证明\( \dfrac{1}{a} + \dfrac{1}{b} < e \)（放缩法）

即证\( x_1 + x_2 < e \)

\( \because f(x_1) = f(x_2) \)，\( \because x_1 \in (0,1) \)，\( \therefore \ln x_1 < 0 \)

\( \therefore x_1(1 – \ln x_1) = x_2(1 – \ln x_2) > x_1 \)

\( \therefore x_2(1 – \ln x_2) > x_1 \)，\( \therefore x_2(2 – \ln x_2) > x_1 + x_2 \)

设\( g(x) = 2x – \ln x \)，定义域\( (1, e) \)

\( g'(x) = 1 – \ln x \)，\( \because x \in (1, e) \)，\( \therefore g'(x) > 0 \)，\( g(x) \)在定义域上单调递增

\( \therefore g(x) < g(e) = e \)，即\( x_1 + x_2 < e \)，即\( \dfrac{1}{a} + \dfrac{1}{b} < e \)