Understanding a solution of a system of differential equations I am looking at the system 
$$
x' = -y + xy^2 \\
y' = 4x - 4x^2 y
$$
and trying to find its solution curves.
I have a proposed solution to this which uses the fact that 
$$
\frac{dy}{dx} = \frac{y'}{x'} = \frac{4x(1-xy)}{y(xy-1)} = -\frac{4x}{y}
$$
and then solves this as a separable equation, which is straight forward enough. What I am struggling to understand is why one chooses to calculate $\frac{dy}{dx}$. It seems to me quite arbitrary to calculate this derivative. What is the connection between this derivative and the original system?
 A: Consider system
$$
\dot x=f(x,y),\\
\dot y=g(x,y),
$$
where $\dot{}$ denotes derivative with respect to $t$. Its solutions are given by $(x(t),y(t))$ and can be represented as curves in 3-dimensional space $\mathbb R\times \mathbb R^2$. However, the system is autonomous, which means that $f$ and $g$ do not depend on $t$ explicitly. In this case instead of dealing with curves in 3D space, we can stick to the plane $(x,y)$, in which the solution curves turns into an orbit $(x(t),y(t))$ where $t$ is now a parameter. Hence an orbit is a projection of the solution curve on the plane $(x,y)$ with a specific time direction (we can put arrows on our orbits).
Finally, how to find actually the orbits. We can solve the original system and then eliminate $t$. Or, differently, we can eliminate $t$ from the original system and consider,
using the fact that
$$
\frac{dx}{dy}=\frac{\frac{dx}{dt}}{\frac{dy}{dt}}=\frac{f(x,y)}{g(x,y)}\,.
$$
Or, similarly, 
$$
\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{g(x,y)}{f(x,y)}\,.
$$
We get a simple equation, but loose the time direction by sticking to this approach. Note that this will not work for those $(x,y)$ where $f(x,y)=g(x,y)=0$, i.e., only for equilibrium points.
