Answer
$y'' = \frac{{2y}}{{{x^2}}}$
Work Step by Step
$$\eqalign{
& {x^3}{y^3} - 4 = 0 \cr
& {x^3}{y^3} = 4 \cr
& {\text{differentiate both sides}} \cr
& \left( {{x^3}{y^3}} \right)' = \left( 4 \right)' \cr
& {\text{product rule}} \cr
& {x^3}\left( {3{y^2}y'} \right) + 3{x^2}{y^3} = 0 \cr
& 3{x^3}{y^2}y' + 3{x^2}{y^3} = 0 \cr
& {\text{find }}y' \cr
& 3{x^3}{y^2}y' = - 3{x^2}{y^3} \cr
& y' = \frac{{ - 3{x^2}{y^3}}}{{3{x^3}{y^2}}} \cr
& y' = - \frac{y}{x} \cr
& {\text{find }}y'' \cr
& y'' = - \left( {\frac{y}{x}} \right)' \cr
& {\text{quotient rule }} \cr
& \left( {\frac{u}{v}} \right)' = \frac{{vu' - uv'}}{{{v^2}}} \cr
& y'' = - \frac{{x\left( y \right)' - y\left( x \right)'}}{{{x^2}}} \cr
& y'' = - \frac{{xy' - y}}{{{x^2}}} \cr
& {\text{replace }}y' = - \frac{y}{x} \cr
& y'' = - \frac{{x\left( { - \frac{y}{x}} \right) - y}}{{{x^2}}} \cr
& {\text{simplify}} \cr
& y'' = - \frac{{ - y - y}}{{{x^2}}} \cr
& y'' = - \frac{{ - 2y}}{{{x^2}}} \cr
& y'' = \frac{{2y}}{{{x^2}}} \cr} $$
