$y(x+y^3)dx=x(y^3-x)dy$ 
Solve $y(x+y^3)dx=x(y^3-x)dy$

Attempt: $$\frac{dy}{dx}=\frac{y(x+y^3)}{x(y^3-x)}.$$
Let $y^3=xt.$ Then $$3y^2\frac{dy}{dx} = t+x\frac{dt}{dx}.$$
Therefore
$$\frac{(t+x\frac{dt}{dx})}{(3y^2)} = \frac{t(t+1)}{y^2(t-1)}$$
and hence
$$x\frac{dt}{dx}=\frac{2t^2+4t}{t-1}.$$
Integrating, $$\ln|cx|=\frac{3}{4}\ln|t+2|-\frac{1}{4}\ln|t|.$$
Simplyfing, $$cx^2=\frac{y^3+2x}{y}$$ but the given answer is $y=cx^{1/3}$.
Where have I made a mistake?

I would like to provide an alternate approach,
Given equation,
$$y(x+y^3)dx=x(y^3-x)dy$$
$$\implies \frac{x+y^3}{y^3-x}=\frac{xdy}{ydx}$$
$$\implies \frac{y^3}{x}=\frac{xdy+ydx}{xdy-ydx}$$
$$\implies y^3 \frac{(xdy-ydx)}{x^2}=\frac{xdy+ydx}{x}$$
$$\implies y^3d(\frac{y}{x})=\frac{d(xy)}{x}$$
$$\implies\frac{y}{x}d(\frac{y}{x})=\frac{d(xy)}{(xy)^2}$$
Integrating, we get
$$\frac{1}{2}\frac{y^2}{x^2}+\frac{1}{xy}=c'$$
If you take $c'=c/2$, it is exactly the same solution you got.
