Changing to polar coordinates bring the differential equation to $\Phi(\phi,\rho,\rho'(\phi)) = 0$ form.

I have a differential equation $y'=\dfrac{x+y}{x-y}$ and the problem says to change to polar coordinate system by assuming $x = \rho\cos\phi$, $y = \rho\sin\phi$ and $\rho=\rho(\phi)$ and bring the equation to $\Phi(\phi,\rho,\rho'(\phi)) = 0$ form. The answer is $\rho'(\phi)=\rho(\phi)$. This problem is under "Implicit Functions" in the book, so I suppose I should use the Implicit-function theorem. But I don't see what I can get from that. I did substitute and tried to simplify, but in vain. The only way I see to solve this after substitution is by applying trigonometry, but every time either $\rho(\phi)$ or $\rho'(\phi)$ are multiplied by some other trigonometric function (that can't be simplified to a constant) and it seems that just trigonometry is not enough. Could you explain how such problems are solved? I couldn't find examples on the internet either.

Multiply out the numerator and re-arrange to get $$ xy'-y=x+yy' $$ or more suggestive $$ x\,dy-y\,dx=x\,dx+y\,dy $$

Under polar coordinates this reads also as $$ ρ^2\,dϕ =ρ\,dρ $$ One can reason that in most situations this can be canceled to $$ \frac{dρ}{dϕ}=ρ $$ with an implied change in the independent variable.

$$y'=\dfrac{x+y}{x-y}$$ $$\dfrac {dy}{dx}=\dfrac{x+y}{x-y}$$ Be carefull and apply the chain rule: $$\dfrac {dy}{dt}\dfrac {dt}{dx}=\dfrac{x+y}{x-y}$$ $$\dfrac {dy}{dt}=\dfrac {dx}{dt}\dfrac{x+y}{x-y}$$ Substitute now $y=r(t)\sin t$: $$\dfrac {dr(t) \sin t}{dt}=\dfrac{\cos t+ \sin t}{\cos t-\sin t}\dfrac {dr(t) \cos(t)}{dt}$$ $$r' \sin t+r \cos t=\dfrac{\cos t+ \sin t}{\cos t-\sin t}(r'\cos t -r \sin t)$$ $$(r' \sin t+r \cos t)(\cos t -\sin t)=(\cos t+ \sin t)(r'\cos t -r \sin t)$$ This is simple trigonometry. It should simplify nicely and give you Lutz Lehmann's nice answer. $$r'=r$$