$x'=x-2y$
$y'=4x-x^3$
Equilibrium points are $(2,1),(-2,-1),(0,0)$
Consider equilibrium point $(2,1)$: Let $X=x-2$ and $Y=y-1$. Subbing this into the main and eliminating all the nonlinear terms gives:
$x'=X-2Y$
$y'=-8X$
Giving the eigenvalues $\lambda_{\pm}=1/2 \pm \sqrt{65}/2$ so $(0,0)$ is a saddle point. The Linearization Theorem implies $(2,1)$ is a saddle point for the nonlinear system. To find the straight line paths in the $XY$-plane, we can either find the eigenvectors $v_{\pm}$ or by subbing $Y=kX$ into the equation for the phase paths $$\frac{dY}{dX}=\frac{-8X}{X-2Y}$$ so $2k^2-k-8=0$ giving $k_{\pm}=1/4 \pm \sqrt{65}/4$, giving the straight line paths $$Y=k_{\pm}X$$ Writing in terms of original terms, we get $$y=k_{\pm}(x-2)+1$$ Now that we got these lines, what exactly do they mean? What is the point of them because I found all of the isoclines and used the fact that the $(0,0)$ is an unstable spiral and found roughly the direction of flow on the phase paths.
How do I know which line corresponds to which eigenvalue??? I really need to know this!!
The eigenvectors tell you the direction of the flow locally around the fixed point. By locally, I mean, very close (how close? as close as you can be) to the fixed point.
Here's a picture:
The pure green and pure blue arrows represent the eigenvectors. The green eigenvector is associated with the negative eigenvalue, and the blue eigenvector is associated with the positive eigenvalue.