Fixed Point of a System Say I have a simple system of the form
$x' = \dfrac{x+1}{y}$
$y' = \dfrac{y+1}{x}$
We know there is at least one stable node fixed point at $(x^{*},y^{*})=(-1,-1)$.  
My question is, is origin also a fixed point (as well as $(x^{*},y^{*})=(-1,0),(0,-1)$)?  
My intuition (which is terrible) is telling me it is not a fixed point since $x'=\dfrac{1}{0}$ and $y'=\dfrac{1}{0}$ for $(x^{*},y^{*})=(0,0)$.  However, whenever I look at a phase portrait I want to say its a fixed point, particularly a saddle point.

For reference, I'm using Strogatz book.
 A: Unfortunately, I do not own any books written by Strogatz. Therefore, I am not familiar with the definitional framework that he uses in his books. I will try to answer the question using a common definitional framework that is used by many authors in the subject (some general references could be "Solving Differential Equations on Manifolds" by Ernst Hairer and "Nonlinear Control Systems" by Horacio J. Marquez). 

Suppose you are given a system of (autonomous) differential equations $Dy=f(y)$, where $f:U\longrightarrow\mathbb{R}^n$, $U$ is an open set such that $U\subseteq \mathbb{R}^n$ (with appropriate smoothness conditions on $f$). An equilibrium point is (sometimes) defined as a function $y$, such that there exists $c \in U$, such that $y(t)=c$ for all $t\in \mathbb{R}$ and $f(c)=0$. 
For your problem, clearly, $n = 2$ and $f(x, y) = (\frac{x+1}{y}, \frac{y+1}{x})^T$ for all $(x, y) \in U$. Therefore, $V \times {\{0\}} \not\subseteq U$ and ${\{0\}} \times W \not\subseteq U$ for all non-empty $V,W\subseteq\mathbb{R}$. Thus, $(0,c)$ and $(c,0)$ are not equilibrium points for all $c$, because they cannot occur in the phase-space. 
For the proposed equilibrium points $(x,y)=(-1,0)$ and $(x,y)=(0,-1)$ there is also another method of reasoning. For example, if $(x,y)=(-1,0)$, then $y'=-1$, which indicates that the system is not in equilibrium. 

A: For $x\ne 0$ and $y\ne0$, you can re-parametrize the system using $t=xys$, denoting $X(s)=x(t)$, $Y(s)=y(t)$ so that
$$
X'(s)=x'(t)xy=X(s)(1+X(s))\\
Y'(s)=y'(t)xy=Y(s)(1+Y(s))\\
$$
This system has the 4 critical points that you got and also the same solution curves as the phase portrait. Note that the direction of the vectors resp. orientation of solution trajectories is reversed in the quadrants with $xy<0$.

As for your original question, in addition to  the answer of xanonec, note that the domain of an ODE is always an open set in phase space. Thus the coordinate axes to not belong to it entirely, even if you could do some kind of continuation in selected points.
