How do I prove these two derivative expressions equivalent?

Question

I was taking the implicit derivative of the following: $$ \frac x y = x - y $$ I'll cut straight to the chase and say my answer was this: $$ \frac {dy}{dx} = \frac {y - y^2} {x - y^2} $$ I've come across something interesting though. If you initially multiply both sides of the equation by $y$, instead differentiating $x = xy - y^2$, you reach the following expression: $$ \frac {dy}{dx} = \frac {1 - y} {x - 2y} $$ I'm sure these two expressions must be equivalent, right? They were both derived from algebraically equivalent expressions, and I confirmed it by brute forcing and plugging x and y values in. My question is this: how can the original expression the derivatives came from be used to make a substitution in terms of x and y in the second case such that it becomes the first case? I intend to understand how to prove they are equivalent via a substitution. Thank you

Answer 1

Substitute $$x-2y=(x-y)-y=\frac{x}y-y=\frac{x-y^2}{y}$$ then multiply the numerator and denominator by $y.$

So you get:

$$\frac{1-y}{x-2y}=\dfrac{1-y}{\frac{x-y^2}y}=\frac{y-y^2}{x-y^2}$$