Answer
$$\eqalign{
& {\text{critical points: }}\left( {0,0} \right),{\text{ }}\left( { - 2,3\root 3 \of {{{\left( { - 2} \right)}^2}} } \right) \cr
& {\text{inflection points: }}\left( {1,6} \right){\text{ and }}\left( {0,0} \right) \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = 5{x^{2/3}} + {x^{5/3}} \cr
& {\text{Differentiate }}f'\left( x \right) \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {5{x^{2/3}} + {x^{5/3}}} \right] \cr
& f'\left( x \right) = \frac{{10}}{3}{x^{ - 1/3}} + \frac{5}{3}{x^{2/3}} \cr
& {\text{Calculate the critical points set the derivative equal to 0}} \cr
& \frac{{10}}{3}{x^{ - 1/3}} + \frac{5}{3}{x^{2/3}} = 0 \cr
& \frac{5}{3}{x^{ - 1/3}}\left( {2 + x} \right) = 0 \cr
& \frac{5}{3}{x^{ - 1/3}} = 0,{\text{ }}2 + x = 0 \cr
& x = - 2 \cr
& {\text{The derivative is not defined at }}x = 0 \cr
& {\text{Calculate }}f\left( 0 \right),{\text{ }}f\left( { - 2} \right) \cr
& f\left( 0 \right) = 0,{\text{ }}f\left( { - 2} \right) = 5{\left( { - 2} \right)^{2/3}} + {\left( { - 2} \right)^{5/3}} = 3\root 3 \of {{{\left( { - 2} \right)}^2}} \cr
& {\text{Therefore, the critical points are:}} \cr
& \left( {0,0} \right),{\text{ }}\left( { - 2,3\root 3 \of {{{\left( { - 2} \right)}^2}} } \right) \cr
& \cr
& {\text{Calculate the second derivative}} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{10}}{3}{x^{ - 1/3}} + \frac{5}{3}{x^{2/3}}} \right] \cr
& f''\left( x \right) = - \frac{{10}}{9}{x^{ - 4/3}} + \frac{{10}}{9}{x^{ - 1/3}} \cr
& \cr
& {\text{Set }}f''\left( x \right) = 0 \cr
& - \frac{{10}}{9}{x^{ - 4/3}} + \frac{{10}}{9}{x^{ - 1/3}} = 0 \cr
& \frac{{10}}{9}{x^{ - 4/3}}\left( { - 1 + x} \right) = 0 \cr
& - 1 + x = 0 \cr
& x = 1 \cr
& {\text{And the second derivative is not defined at }}x = 0 \cr
& {\text{Calculate }}f\left( 1 \right){\text{ and }}f\left( 0 \right) \cr
& f\left( 1 \right) = 6 \cr
& f\left( 0 \right) = 0 \cr
& {\text{Therefore, the inflection points are:}} \cr
& \left( {1,6} \right){\text{ and }}\left( {0,0} \right) \cr} $$
