Answer
$$x = 0,{\text{ }}x = \root 3 \of {16} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{{x^2}}}{{{x^3} + 8}} \cr
& {\text{Calculate the derivative of the function}} \cr
& f'\left( x \right) = \frac{{\left( {{x^3} + 8} \right)\frac{d}{{dx}}\left[ {{x^2}} \right] - {x^2}\frac{d}{{dx}}\left[ {{x^3} + 8} \right]}}{{{{\left( {{x^3} + 8} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{\left( {{x^3} + 8} \right)\left( {2x} \right) - {x^2}\left( {3{x^2}} \right)}}{{{{\left( {{x^3} + 8} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{2{x^4} + 16x - 3{x^4}}}{{{{\left( {{x^3} + 8} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{ - {x^4} + 16x}}{{{{\left( {{x^3} + 8} \right)}^2}}} \cr
& {\text{Factor the numerator}} \cr
& f'\left( x \right) = \frac{{ - x\left( {{x^3} - 16} \right)}}{{{{\left( {{x^3} + 8} \right)}^2}}} \cr
& {\text{Set }}f'\left( x \right) = 0 \cr
& - x\left( {{x^3} - 16} \right) = 0 \cr
& - x\left( {{x^3} - 16} \right) = 0 \cr
& - x = 0,{\text{ }}{x^3} - 16 = 0 \cr
& x = 0,{\text{ }}{x^3} = 16 \cr
& x = 0,{\text{ }}x = \root 3 \of {16} \cr
& {\text{Therefore, the stationary points are:}} \cr
& x = 0,{\text{ }}x = \root 3 \of {16} \cr} $$
