Answer
$${\text{relative minimum of 0 at }}x = 0$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{{x^2}}}{{{x^2} + 1}} \cr
& {\text{Calculate the first derivative }} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^2}}}{{{x^2} + 1}}} \right] \cr
& f'\left( x \right) = \frac{{2x\left( {{x^2} + 1} \right) - {x^2}\left( {2x} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{2{x^3} + 2x - 2{x^3}}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{2x}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr
& {\text{Find the critical point}}{\text{, set }}f'\left( x \right) = 0 \cr
& \frac{{2x}}{{{{\left( {{x^2} + 1} \right)}^2}}} = 0 \cr
& {\text{The critical points is }}x = 0 \cr
& \cr
& {\text{Calculate the second derivative }} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right] \cr
& f''\left( x \right) = \frac{{2{{\left( {{x^2} + 1} \right)}^2} - 4x\left( {{x^2} + 1} \right)\left( {2x} \right)}}{{{{\left( {{x^2} + 1} \right)}^4}}} \cr
& f''\left( x \right) = \frac{{2\left( {{x^2} + 1} \right) - 4x\left( {2x} \right)}}{{{{\left( {{x^2} + 1} \right)}^3}}} \cr
& f''\left( x \right) = \frac{{2{x^2} + 2 - 8{x^2}}}{{{{\left( {{x^2} + 1} \right)}^3}}} \cr
& f''\left( x \right) = \frac{{2 - 6{x^2}}}{{{{\left( {{x^2} + 1} \right)}^3}}} \cr
& {\text{Using the second test derivative into the critical points}} \cr
& {\text{Evaluate }}f''\left( 0 \right) \cr
& f''\left( 0 \right) = \frac{{2 - 6{{\left( 0 \right)}^2}}}{{{{\left( {{{\left( 0 \right)}^2} + 1} \right)}^3}}} \cr
& f''\left( 0 \right) = 2 \cr
& f''\left( 0 \right) > 0,{\text{ then }}f\left( x \right){\text{ has a relative minimum at }}x = 0 \cr} $$