Solving the ODE $(x^2 y-1)y'-(xy^2-1)=0$ I need to solve $(x^2 y-1)y'-(x y^2-1)=0$.
Not sure how to approach this ODE, would love some help regarding solving it or any useful resources.

Too long for a comment
Using @projectilemotion's answer, we can easily solve the cubic equation
$$y^3-3x y^2+\frac{3 (2 C+1) x^2}{2 C}y-\frac {(2 C+1) x^3+2 }{2C}=0$$
For $C=1$, the solution is
$$y=x \left(1-\sqrt{2} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\sqrt{2}\frac{
   \left(x^3-1\right)}{x^3}\right)\right)\right)$$
For $C=2$,
$$y=x \left(1-\sinh \left(\frac{1}{3} \sinh ^{-1}\left(2-\frac{2}{x^3}\right)\right)\right)$$
So, trying the substitution $y=x \left(1-e^{z}\right)$, the differential equation becomes
$$\left(x^4 \left(e^{z}-1\right)+x\right) z'+1=0$$ Switch variables
$$\frac{\left(e^z-1\right) x^4+x}{x'}+1=0$$ Let $x=\frac 1{\sqrt[3]t}$
$$t'-3 t-3 e^z+3=0\quad \implies t=c_1 \,e^{3 z}+\frac{1}{2} \left(2-3 e^z\right)$$
Now, back to $x$ and $y$ for the implicit solution.
