Answer
$$\frac{dy}{dx}=\frac{7y-3x^2(x+3y)^2}{7x}$$
Work Step by Step
$$x^3=\frac{2x-y}{x+3y}$$
1) Differentiate both sides of the equation with respect to $x$:
$$\frac{d}{dx}(x^3)=\frac{d}{dx}\Big(\frac{2x-y}{x+3y}\Big)=\frac{(2-\frac{dy}{dx})(x+3y)-(2x-y)(1+3\frac{dy}{dx})}{(x+3y)^2}$$ $$3x^2=\frac{(2x+6y)-\frac{dy}{dx}(x+3y)-(2x-y)-\frac{dy}{dx}(6x-3y)}{(x+3y)^2}$$ $$3x^2=\frac{7y-\frac{dy}{dx}(x+3y)-\frac{dy}{dx}(6x-3y)}{(x+3y)^2}$$ $$7y-\frac{dy}{dx}(x+3y)-\frac{dy}{dx}(6x-3y)=3x^2(x+3y)^2$$
2) Collect all the terms with $dy/dx$ on one side and solve for $dy/dx$: $$\frac{dy}{dx}(x+3y)+\frac{dy}{dx}(6x-3y)=7y-3x^2(x+3y)^2$$ $$7x\frac{dy}{dx}=7y-3x^2(x+3y)^2$$ $$\frac{dy}{dx}=\frac{7y-3x^2(x+3y)^2}{7x}$$
