Answer
$$\frac{{dy}}{{dx}} = \frac{{ - 3x{{\left( {2 + y} \right)}^2}}}{2}$$
Work Step by Step
$$\eqalign{
& 3{x^2} = \frac{{2 - y}}{{2 + y}} \cr
& {\text{take the derivative on both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {3{x^2}} \right) = \frac{d}{{dx}}\left( {\frac{{2 - y}}{{2 + y}}} \right) \cr
& {\text{use quotient rule for derivatives on the right side}} \cr
& \frac{d}{{dx}}\left( {3{x^2}} \right) = \frac{{\left( {2 + y} \right)d/dx\left( {2 - y} \right) - \left( {2 - y} \right)d/dx\left( {2 + y} \right)}}{{{{\left( {2 + y} \right)}^2}}} \cr
& {\text{solve the derivatives }} \cr
& 6x = \frac{{\left( {2 + y} \right)\left( { - \frac{{dy}}{{dx}}} \right) - \left( {2 - y} \right)\left( {\frac{{dy}}{{dx}}} \right)}}{{{{\left( {2 + y} \right)}^2}}} \cr
& 6x{\left( {2 + y} \right)^2} = \left( {2 + y} \right)\left( { - \frac{{dy}}{{dx}}} \right) - \left( {2 - y} \right)\left( {\frac{{dy}}{{dx}}} \right) \cr
& {\text{combine terms and solve for }}\frac{{dy}}{{dx}} \cr
& 6x{\left( {2 + y} \right)^2} = \left[ { - \left( {2 + y} \right) - \left( {2 - y} \right)} \right]\frac{{dy}}{{dx}} \cr
& 6x{\left( {2 + y} \right)^2} = \left( { - 2 - y - 2 + y} \right)\frac{{dy}}{{dx}} \cr
& 6x{\left( {2 + y} \right)^2} = \left( { - 4} \right)\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{{6x{{\left( {2 + y} \right)}^2}}}{{ - 4}} \cr
& \frac{{dy}}{{dx}} = \frac{{ - 3x{{\left( {2 + y} \right)}^2}}}{2} \cr} $$
