Concept:         

Second Derivative Test:

• Calculate f’(x)
• Solve f’(x) = 0 and find the roots. Suppose x = c is the root of f’(x) = 0.
• Calculate f’’(x) and put x = c to get the value of f’’(c).

• If f’’(c) < 0 then x = c is a point of maxima.
• If f’’(c) > 0 then x = c is a point of minima.
• If f’’(c) = 0 then we need to use the first derivative test.

Calculation:      

Given: \(f(x)={(x^2-x+1) \over(x^2+x+1) } \)

\(\frac{{df}}{{dx}} = \frac{{({x^2} + x + 1)(2x - 1) - ({x^2} - x + 1)(2x + 1)}}{{{{({x^2} + x + 1)}^2}}}\)

\(\frac{{df}}{{dx}} = \frac{{{x^2} - 1}}{{{{({x^2} + x + 1)}^2}}}\)

\(\frac{{df}}{{dx}} = 0 ⇒ {x^2} - 1 = 0 ⇒ x = \pm 1\)

\(⇒ \frac{{{d^2}f}}{{d{x^2}}} = \frac{{\left( {2x - 1} \right)\left( {{x^2} + x + 1} \right) - 2\left( {{x^2} - 1} \right)\left( {2x + 1} \right)}}{{{{\left( {{x^2} + x + 1} \right)}^3}}}\)

⇒ f''(1) = 1/9 > 0

So, x = 1 is  a point of minima

⇒ f''(-1) = - 1/9 < 0

So, x = - 1 is a point of maxima

So minimum value of the given function is attained at x = 1

By substituting x = 1 in f(x) we get f(1) = 1/3.

Hence, option D is the correct answer.