Answer
$\frac{{{d^2}y}}{{d{x^2}}} = - \frac{{2x}}{{{y^5}}}$
Work Step by Step
$$\eqalign{
& {x^3} + {y^3} = 1 \cr
& {\text{differentiate both sides with respect to }}x \cr
& \left( {{x^3} + {y^3}} \right)' = \left( 1 \right)' \cr
& \left( {{x^3}} \right)' + \left( {{y^3}} \right)' = \left( 1 \right)' \cr
& 3{x^2} + 3{y^2}y' = 0 \cr
& {\text{find }}y'{\text{ subtract }}3{x^2}{\text{ from each side}} \cr
& 3{y^2}y' = - 3{x^2} \cr
& y' = \frac{{ - 3{x^2}}}{{3{y^2}}} \cr
& {\text{cancel common factors}} \cr
& y' = - \frac{{{x^2}}}{{{y^2}}} \cr
& {\text{find }}y'' \cr
& y'' = \frac{d}{{dx}}\left( {y'} \right) \cr
& y'' = - \left( {\frac{{{x^2}}}{{{y^2}}}} \right)' \cr
& {\text{quotient rule }} \cr
& \left( {\frac{u}{v}} \right)' = \frac{{vu' - uv'}}{{{v^2}}} \cr
& y'' = - \frac{{{y^2}\left( {{x^2}} \right)' - {x^2}\left( {{y^2}} \right)'}}{{{y^4}}} \cr
& y'' = - \frac{{2x{y^2} - 2{x^2}yy'}}{{{y^4}}} \cr
& {\text{replace }}y' = - \frac{{{x^2}}}{{{y^2}}} \cr
& y'' = - \frac{{2x{y^2} - 2{x^2}y\left( { - \frac{{{x^2}}}{{{y^2}}}} \right)}}{{{y^4}}} \cr
& {\text{simplify}} \cr
& y'' = - \frac{{2x{y^2} + \frac{{2{x^4}}}{y}}}{{{y^4}}} \cr
& y'' = - \frac{{2x{y^3} + 2{x^4}}}{{{y^5}}} \cr
& {\text{factor}} \cr
& y'' = - \frac{{2x\left( {{y^3} + {x^3}} \right)}}{{{y^5}}} \cr
& {\text{We know that }}{x^3} + {y^3} = 1 \cr
& y'' = - \frac{{2x\left( 1 \right)}}{{{y^5}}} \cr
& y'' = - \frac{{2x}}{{{y^5}}} \cr
& {\text{we can write }}y''{\text{ as }}\frac{{{d^2}y}}{{d{x^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{2x}}{{{y^5}}} \cr} $$
