Answer
$$\frac{{{d^2}y}}{{d{x^2}}} = - \frac{8}{{9{y^3}}}$$
Work Step by Step
$$\eqalign{
& 2{x^2} - 3{y^2} = 4 \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ {2{x^2} - 3{y^2}} \right] = \frac{d}{{dx}}\left[ 4 \right] \cr
& 4x - 6y\frac{{dy}}{{dx}} = 0 \cr
& \frac{{dy}}{{dx}} = \frac{{4x}}{{6y}} \cr
& \frac{{dy}}{{dx}} = \frac{{2x}}{{3y}} \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{3y\left( 2 \right) - 2x\left( 3 \right)\frac{{dy}}{{dx}}}}{{{{\left( {3y} \right)}^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{6y - 6x\frac{{dy}}{{dx}}}}{{9{y^2}}} \cr
& {\text{Substituting }}\frac{{dy}}{{dx}} = \frac{{2x}}{{3y}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{6y - 6x\left( {\frac{{2x}}{{3y}}} \right)}}{{9{y^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{6y - \frac{{4{x^2}}}{y}}}{{9{y^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{6{y^2} - 4{x^2}}}{{9{y^3}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - 2\left( {2{x^2} - 3{y^2}} \right)}}{{9{y^3}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - 2\left( 4 \right)}}{{9{y^3}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - \frac{8}{{9{y^3}}} \cr} $$
