Can a non-constant solution of DE intersect an equilibrium solution. 
Consider the autonomous DE;
  $$\frac{dy}{dx}=\frac{y^3-9y}{e^y}$$
  Find the equillibrium solutions and classify each as stable, unstable or semi-stable. Can a non-constant solution of this differential equation intersect an equillibrium solution? Justify why or why not.

I am have issues with the second part, but here is all of my work.
My Attempt:
$$\frac{dy}{dx}=0=\frac{y(y+3)(y-3)}{e^y} \implies$$
Equillibrium solutions are $y=0,-3,3$
From what I get;
$$y=-3, \text{unstable}$$
$$y=0, \text{stable}$$
$$y=3, \text{unstable}$$
Through separation of variables I get;
$$\frac{e^y}{y^3-9y}dy=dx$$
Assuming I am correct up to here, do I need to integrate to answer the question?(If so how can I do that? It looks like a very challenging integral.) Is it based on the points being stable or unstable? I am not sure how to answer;

"Can a non-constant solution of this differential equation intersect an equillibrium solution? Justify why or why not."

Thanks in advance!
 A: The answer is no because the equation is "nice enough" that its initial value problems have unique solutions. If a non-constant solution could intersect an equilibrium solution, then there would be multiple solutions to an IVP where $y_0$ is equal to that equilibrium value.
An example of an equation lacking this property is $y'=|y|^{1/2}$, for which there are multiple solutions to the IVP whenever $y_0 \leq 0$.
A: You don't need to integrate $\frac{dy}{dx}=\frac{y^3-9y}{e^y}$ if you know the theorem about existence and uniqueness of the solutions to initial problem and if you are allowed to use it.
If the answer is required without applying this theorem, you can proceed this way :
$$x=\int \frac{e^y}{y^3-9y}dy$$
$$x(y)=-\frac19\int\frac{e^y}{y}dy+\frac{1}{18}\int\frac{e^y}{y+3}dy+\frac{1}{18}\int\frac{e^y}{y-3}dy+C$$
Note : These integrals are related to a special function namely "Exponential Integrals", http://mathworld.wolfram.com/ExponentialIntegral.html , but we don't need to refer to it.
Obviously the function $x(y)$ is finite only in ranges of $y$ which do not include $y=0$ , $y=-3$ and $y=3$, that is on the ranges $y<-3$ , $-3<y<0$ , $0<y<3$ and $y>3$.
In other words, if a non-constant solution would exist, $x$ would be infinite at $y=0$ , $y=-3$ and $y=3$, so could not go through the equilibrium points at finite values of $x$. Thus the answer is no.
