{dx \over dt}=P(x,y), \qquad {dy \over dt}=Q(x,y)

$${dx \over dt}=P(x,y), \qquad {dy \over dt}=Q(x,y)$$

 

do you want   $$\frac{dy}{dx}?$$

 

Well         $$\\{dx \over dt}={P(x,y)\over 1}\qquad \mbox{therefore}\\\\
{dt \over dx}={1 \over P(x,y)} \\\\
so\\\\
\dfrac{dy}{dt}\times\dfrac{dt}{dx}=Q(x,y)\times \dfrac{1}{P(x,y)}\\\\\\
\dfrac{dy}{dx}=\dfrac{Q(x,y)}{P(x,y)}$$

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