Answer
$$f''\left( x \right) = - \frac{1}{{{x^2}}} + \frac{2}{{{x^3}}},\,\,\,\,\,f''\left( 0 \right) = {\text{undefined}},\,\,\,\,\,f''\left( 2 \right) = 0$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \ln x + \frac{1}{x} \cr
& {\text{write the function as}} \cr
& f\left( x \right) = \ln x + {x^{ - 1}} \cr
& {\text{find the derivative of }}f\left( x \right),{\text{ use }}\frac{d}{{dx}}\left[ {\ln x} \right] = \frac{1}{x}{\text{ and }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}}.{\text{ then}} \cr
& f'\left( x \right) = \frac{1}{x} - {x^{ - 2}} \cr
& or \cr
& f'\left( x \right) = {x^{ - 1}} - {x^{ - 2}} \cr
& \cr
& {\text{find the derivative of }}f'\left( x \right) \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {{x^{ - 1}} - {x^{ - 2}}} \right] \cr
& {\text{use }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}}.{\text{ then}} \cr
& f''\left( x \right) = - {x^{ - 2}} - \left( { - 2{x^{ - 3}}} \right) \cr
& f''\left( x \right) = - {x^{ - 2}} + 2{x^{ - 3}} \cr
& or \cr
& f''\left( x \right) = - \frac{1}{{{x^2}}} + \frac{2}{{{x^3}}} \cr
& \cr
& {\text{find }}f''\left( 0 \right){\text{ and }}f''\left( 2 \right) \cr
& f''\left( 0 \right) = - \frac{1}{{{{\left( 0 \right)}^2}}} + \frac{2}{{{{\left( 0 \right)}^3}}} \cr
& {\text{then the second derivative is not defined for }}x = 0 \cr
& f''\left( 2 \right) = - \frac{1}{{{{\left( 2 \right)}^2}}} + \frac{2}{{{{\left( 2 \right)}^3}}} = - \frac{1}{4} + \frac{1}{4} = 0 \cr} $$
