Proving an implicit equation has no orbits

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Suppose we have the function $F(x,y)=x+y-a\log(x)$ where a is a constant and define the orbit by the implicit equation $F(x,y)=c$ where c is a constant. How can I formally prove that no orbit reaches the $y$-axis?

I can sketch the some level curves of this, but still I want to know if this is something that can be proven formally?

Thanks in advance!

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$F(x,y)$ is not defined when $x=0$. So if an "orbit" is defined as the set of solutions of $F(x,y)=c$, it can't contain any point of the $y$ axis.

EDIT: Ah, so what's really going on here is that $F(x,y) = c$ is an implicit form for the trajectories of a system of differential equations: those are the "orbits". You should have mentioned that in the first place.

Well, not only does the solution set of $F(x,y) = c$ not contain any point of the $y$ axis, it also has no limit points on the $y$ axis if $a\ne 0$, because $\lim_{x \to 0+} \log(x) = -\infty$.