Answer
$${\text{ }}f\left( x \right){\text{ has a relative maximum at }}x = - \frac{{27}}{{64}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 2x + {x^{2/3}} \cr
& {\text{Calculate the derivative of }}f\left( x \right) \cr
& f'\left( x \right) = 2 + \frac{3}{2}{x^{ - 1/3}} \cr
& {\text{Find the critical points, set }}f'\left( x \right) = 0 \cr
& 2 + \frac{3}{2}{x^{ - 1/3}} = 0 \cr
& 2 + \frac{3}{{2\root 3 \of x }} = 0 \cr
& \frac{3}{{2\root 3 \of x }} = - 2 \cr
& \frac{3}{2} = - 2\root 3 \of x \cr
& \root 3 \of x = - \frac{3}{4} \cr
& x = {\left( { - \frac{3}{4}} \right)^3} \cr
& x = - \frac{{27}}{{64}} \cr
& {\text{Calculate the second derivative}} \cr
& f''\left( x \right) = \frac{3}{2}\left( { - \frac{1}{3}} \right){x^{ - 4/3}} \cr
& f''\left( x \right) = - \frac{1}{2}{x^{ - 4/3}} \cr
& {\text{Use the second test derivative into }}x = - \frac{{27}}{{64}} \cr
& f''\left( { - \frac{{27}}{{64}}} \right) = - \frac{1}{2}{\left( { - \frac{{27}}{{64}}} \right)^{ - 4/3}} \cr
& f''\left( { - \frac{{27}}{{64}}} \right) = - \frac{{128}}{{81}} \cr
& f''\left( { - \frac{{27}}{{64}}} \right) < 0,{\text{ then }}f\left( x \right){\text{ has a relative maximum at }}x = - \frac{{27}}{{64}} \cr} $$