Answer
$$\frac{{dy}}{{dx}} = - \frac{{{x^{ - 1/3}}}}{{{y^{ - 1/3}}}}{\text{ and }}\frac{{{d^2}y}}{{d{x^2}}} = \frac{1}{3}{x^{ - 4/3}}{y^{1/3}} + \frac{1}{3}{x^{ - 2/3}}{y^{ - 1/3}}$$
Work Step by Step
$$\eqalign{
& {x^{2/3}} + {y^{2/3}} = 1 \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {{x^{2/3}}} \right) + \frac{d}{{dx}}\left( {{y^{2/3}}} \right) = \frac{d}{{dx}}\left( 1 \right) \cr
& \cr
& {\text{find derivatives}} \cr
& \frac{2}{3}{x^{ - 1/3}} + \frac{2}{3}{y^{ - 1/3}}\frac{{dy}}{{dx}} = 0 \cr
& \cr
& {\text{solve for }}\frac{{dy}}{{dx}} \cr
& \frac{2}{3}{y^{ - 1/3}}\frac{{dy}}{{dx}} = - \frac{2}{3}{x^{ - 1/3}} \cr
& \frac{{dy}}{{dx}} = \frac{{ - \frac{2}{3}{x^{ - 1/3}}}}{{\frac{2}{3}{y^{ - 1/3}}}} \cr
& \frac{{dy}}{{dx}} = - \frac{{{x^{ - 1/3}}}}{{{y^{ - 1/3}}}} \cr
& \cr
& {\text{find the second derivative}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( { - \frac{{{x^{ - 1/3}}}}{{{y^{ - 1/3}}}}} \right) \cr
& {\text{use the quotient rule}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{{y^{ - 1/3}}\frac{d}{{dx}}\left( {{x^{ - 1/3}}} \right) - {x^{ - 1/3}}\frac{d}{{dx}}\left( {{y^{ - 1/3}}} \right)}}{{{y^{ - 2/3}}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{{y^{ - 1/3}}\left( { - \frac{1}{3}{x^{ - 4/3}}} \right) - {x^{ - 1/3}}\left( { - \frac{1}{3}{y^{ - 4/3}}} \right)\frac{{dy}}{{dx}}}}{{{y^{ - 2/3}}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{1}{3}{x^{ - 4/3}}{y^{ - 1/3}} - \frac{1}{3}{x^{ - 1/3}}{y^{ - 4/3}}\frac{{dy}}{{dx}}}}{{{y^{ - 2/3}}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{1}{3}{x^{ - 4/3}}{y^{1/3}} - \frac{1}{3}{x^{ - 1/3}}{y^{ - 2/3}}\frac{{dy}}{{dx}} \cr
& {\text{substitute }}\frac{{dy}}{{dx}} = - \frac{{{x^{ - 1/3}}}}{{{y^{ - 1/3}}}}{\text{ and simplify}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{1}{3}{x^{ - 4/3}}{y^{1/3}} - \frac{1}{3}{x^{ - 1/3}}{y^{ - 2/3}}\left( { - \frac{{{x^{ - 1/3}}}}{{{y^{ - 1/3}}}}} \right) \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{1}{3}{x^{ - 4/3}}{y^{1/3}} + \frac{1}{3}{x^{ - 2/3}}{y^{ - 1/3}} \cr} $$
