Answer
$$\frac{{dy}}{{dx}} = - \frac{{4x + 3y}}{{3x - 12{y^3}}}$$
Work Step by Step
$$\eqalign{
& 2{x^2} + 3xy - 3{y^4} = 2 \cr
& {\text{Let }}F\left( {x,y} \right) = 2{x^2} + 3xy - 3{y^4} - 2 \cr
& {\text{Calculate the partial derivatives }}{F_x}\left( {x,y} \right){\text{ and }}{F_y}\left( {x,y} \right) \cr
& {F_x}\left( {x,y} \right) = 4x + 3y \cr
& {F_y}\left( {x,y} \right) = 3x - 12{y^3} \cr
& {\text{Use the implicit differentiation }}\frac{{dy}}{{dx}} = - \frac{{{F_x}}}{{{F_y}}} \cr
& \frac{{dy}}{{dx}} = - \frac{{4x + 3y}}{{3x - 12{y^3}}} \cr} $$
