exercise: ordinary differential equations I am struggling with an exercise. Can you please give me a hint?
Exercise:
Show that the solution curves of the differential equation:
$\frac{dy}{dx}=-\frac{y(2x^3-y^3)}{x(2y^3-x^3)}$, are of the form $x^3+y^3=3Cxy$.
I tried the substitution $u=y/x \rightarrow y=xu, \frac{dy}{dx}=u+x\frac{du}{dx}$.
Hence I get:
$u+x\frac{du}{dx}=-\frac{u(2-u^3)}{2u^3-1}$
This gives:
$2u^4-u+(2u^3-1)x\frac{du}{dx}=-2u+u^4$
$(2u^3-1)x\frac{du}{dx}=-(u^4+u)$
$\frac{(2u^3-1)}{(u^4+u)}\frac{du}{dx}=-\frac{1}{x}$
So I can atleast reduce the problem to a seperable differential equation, but I am not able to integrate the left side up. Do you have any tips?

Use partial fractions
$$\begin{align}
\frac{2u^{3} - 1}{u^{4} + u} &= \frac{2u^{3} - 1}{u(u^{3} + 1)} \\
&= \frac{A}{u} + \frac{Bu^{2} + Cu + D}{u^{3} + 1} \\
\implies 2u^{3} - 1 &= (A + B)u^{3} + Cu^{2} + Du + A \\
\end{align}$$
Equating coefficients
$$\begin{align}
A &= -1 \\
B &= 3 \\
\end{align}$$
Hence
$$\begin{align}
\frac{2u^{3} - 1}{u^{4} + u} &= \frac{3u^{2}}{u^{3} + 1} - \frac{1}{u}
\end{align}$$
Integrating
$$\begin{align}
\int \bigg(\frac{3u^{2}}{u^{3} + 1} - \frac{1}{u} \bigg)du &= - \int \frac{1}{x} dx \\
\implies \ln(u^{3} + 1) - \ln(u) &= -\ln(x) + K_1 \\
\implies \ln \bigg( \frac{u^{3} + 1}{u} \bigg) &= - \ln(x) + K_1 \\
\implies \frac{u^{3} + 1}{u} &= \frac{K_2}{x} \\
\end{align}$$
With 
$$\begin{align}
u &= \frac{y}{x} \\
\implies \frac{y^{2}}{x^{2}} + \frac{x}{y} &= \frac{K_2}{x} \\
\implies y^{3} + x^{3} &= K_2xy
\end{align}$$
