<s>[INST] 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?
 [/INST] here is a cheap way of doing this knowing that $x^3 + y^3 = Cxy$ is a solution. we will make the change of variables $$u = x^3+y^3, t = xy$$ 
the deifferential equation satisfied by $u$ and $t$ are:
$$\frac{du}{dx} = 3x^2 + 3y^2 \frac{dy}{dx} = \frac{3(y^6-x^6)}{x(2y^3 - x^3}, \frac{dt}{dx} = x\frac{dy}{dx} + y = \frac{3y(y^3-x^3)}{2y^3 - x^3}  $$ 
we can divide it out and get a separable equation $$\frac{du}{dt} = \frac{u}{t} $$ which has solution of the form $$u = Ct, x^3 + y^3 = Cxy$$
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