Answer
$$\frac{{dy}}{{dx}} = \frac{{7y - 3{x^2}{{\left( {x + 3y} \right)}^2}}}{{7x}}$$
Work Step by Step
$$\eqalign{
& {x^3} = \frac{{2x - y}}{{x + 3y}} \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {{x^3}} \right) = \frac{d}{{dx}}\left( {\frac{{2x - y}}{{x + 3y}}} \right) \cr
& {\text{use the quotient rule on the right side}} \cr
& \frac{d}{{dx}}\left( {{x^3}} \right) = \frac{{\left( {x + 3y} \right)\frac{d}{{dx}}\left( {2x - y} \right) - \left( {2x - y} \right)\frac{d}{{dx}}\left( {x + 3y} \right)}}{{{{\left( {x + 3y} \right)}^2}}} \cr
& {\text{solve derivatives}} \cr
& 3{x^2} = \frac{{\left( {x + 3y} \right)\left( {2 - \frac{{dy}}{{dx}}} \right) - \left( {2x - y} \right)\left( {1 + 3\frac{{dy}}{{dx}}} \right)}}{{{{\left( {x + 3y} \right)}^2}}} \cr
& {\text{simplify}} \cr
& 3{x^2} = \frac{{2x - x\frac{{dy}}{{dx}} + 6y - 3y\frac{{dy}}{{dx}} - 2x - 6x\frac{{dy}}{{dx}} + y + 3y\frac{{dy}}{{dx}}}}{{{{\left( {x + 3y} \right)}^2}}} \cr
& 3{x^2} = \frac{{ - 7x\frac{{dy}}{{dx}} + 7y}}{{{{\left( {x + 3y} \right)}^2}}} \cr
& {\text{solve for }}\frac{{dy}}{{dx}} \cr
& 3{x^2}{\left( {x + 3y} \right)^2} = - 7x\frac{{dy}}{{dx}} + 7y \cr
& 3{x^2}{\left( {x + 3y} \right)^2} - 7y = - 7x\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{{7y - 3{x^2}{{\left( {x + 3y} \right)}^2}}}{{7x}} \cr} $$
