Originally Posted by
Soroban Hello, mbempeni!
The first one is homegeneous . . .
Divide top and bottom of the fraction by $\displaystyle x^3\!:\quad\frac{dy}{dx} \;=\;\frac{1 - 2\left(\dfrac{y}{x}\right) - \left(\dfrac{y}{x}\right)^3}{1 - \left(\dfrac{y}{x}\right)^2} $
Let $\displaystyle v \,=\,\frac{y}{x}\quad\Rightarrow\quad y \,=\,vx\quad\Rightarrow\quad \frac{dy}{dx} \:=\:v + x\frac{dv}{dx}$
Substitute: .$\displaystyle v + x\frac{dv}{dx} \;=\;\frac{1-2v-v^3}{1-v^2}$
Then: .$\displaystyle x\frac{dv}{dx} \;=\;\frac{1-2v-v^3}{1-v^2} - v \quad\Rightarrow\quad x\frac{dv}{dx}\;=\;\frac{1-3v}{1-v^2}$
Separate variables: .$\displaystyle \frac{v^2-1}{3v-1}\,dv \;=\;\frac{dx}{x}$
Can you finish it now?