The following example is from the book of Hubbard on Vector Calculus, Linear Algebra and Differential Forms.
It considers the vector field $F(x,y)=(xy-2, x-y)$, and describes pictorially as below.
From the picture, we can see two things, and they raised question in my mind, which I was unable to find in any explanation in the book.
(1) In the third quadrant of $X-Y$ coordinate plane, the vectors of vector field seem circulating around a point. My question is, what it represents, and how do we determine such point from mathematical description of vector field $F$?
(2) In first quadrant of $X-Y$ coordinate plane, the vectors of vector field are avoiding some point; so second question is naturally, what it represents and how do we determine such point from mathematical description of $F$?
(1) and (2) are so-called stationary points. That means, the vector field is 0, there, i.e., $$ F(x,y) = (0,0) $$
If you plug in $ (x,y) = (-\sqrt{2}, -\sqrt{2}) $ (this is the case (1) in the third quadrant), then $F(x,y) = ((-\sqrt{2})^2-2, -\sqrt{2}+\sqrt{2}) = (0,0)$, so the point is indeed stationary.
The same happens for (2) in the first quadrant: for $(x,y)=(\sqrt{2},\sqrt{2})$, you get $F(x,y) = (\sqrt{2}^2-2, \sqrt{2}-\sqrt{2}) = (0,0)$
You may determine these stationary points by solving the system of equations given by $F(x,y)=0$. In our case, it has two variables and two equations: \begin{align} xy-2 &= 0\\ x-y &= 0\\ \end{align} From the second line, you get $ x=y $. Plugging this into the first line gives you $ x^2-2 = 0 \; \Leftrightarrow \; x = \pm \sqrt{2}$. So you have the two solutions $ (x,y)=(\sqrt{2},\sqrt{2}) $ and $ (x,y)=(-\sqrt{2},-\sqrt{2}) $.
What is a bit more difficult is to see whether the vectors are pointing towards or away from a critical point (linear stability analysis). For this, you need to compute the Jacobian matrix: If we write $ F(x,y) = (F_1(x,y),F_2(x,y)) $, as in our case $ F_1(x,y) = xy-2 $, $ F_2(x,y) = x-y $, then the Jacobian is $$DF(x,y) = \begin{pmatrix} \partial_x F_1 & \partial_y F_1 \\ \partial_x F_2 & \partial_y F_2 \end{pmatrix} = \begin{pmatrix} y & x \\ 1 & -1 \end{pmatrix} $$ For instance at the stationary points $ (\sqrt{2},\sqrt{2}) $ and $ (\sqrt{2},\sqrt{2}) $ it evaluates to $$DF(\sqrt{2},\sqrt{2}) = \begin{pmatrix} \sqrt{2} & \sqrt{2} \\ 1 & -1 \end{pmatrix} \qquad \text{and} \qquad DF(-\sqrt{2},-\sqrt{2}) = \begin{pmatrix} -\sqrt{2} & -\sqrt{2} \\ 1 & -1 \end{pmatrix} $$ You now compute the eigenvalues (which are in $ \mathbb{C} $ and which might take some minutes if done by hand) and consider their real parts:
Two eigenvalues with positive real part show to you that the vector field points away from the stationary point, so the point "repels" field lines.
Two eigenvalues with negative real part show to you that the vector field points towards the stationary point, so the point "attracts" field lines.
If one eigenvalue is positive and one negative then some lines point towards and some away from the stationary point, so you have neither an attractive nor a repulsive point.
An eigenvalue 0 doesn't tell you much about the stability of a point - you would need to consider further derivatives or plot the vector field i order to find out, whether field lines point towards or away from it.