How do I linearise the rational function to analyze the critical points? 
For the system $$\frac{dx}{dt}=\frac{3xy}{1+x^2+y^2}-\frac{1+x^2}{1+y^2}\\\frac{dy}{dt}=x^2-y^2,$$ the point $\begin{pmatrix}1\\1\end{pmatrix}$ is
A. an unstable node
B. a stable node
C. a saddle point
D. a stable spiral point
E. an unstable spiral point.

I understand I need to shift the point to the origin and just eliminate $x^2$ and $y^2$, but how do I deal with denominators?
Thank you
 A: This is a nonlinear system, so we can analyze behavior at critical points by finding the Jacobian matrix evaluated at those points and then computing eigenvalues of that matrix. Recall, the Jacobian matrix is:
$$\begin{bmatrix}\frac{\partial f}{\partial x}&\frac{\partial f}{\partial y}\\\frac{\partial g}{\partial x}&\frac{\partial g}{\partial y}\end{bmatrix}$$
Where $f(x,y)=\frac{dx}{dt}$ and $g(x,y)=\frac{dy}{dt}$.
So we have the following Jacobian matrix:
$$\begin{bmatrix}\frac{3y(1-x^2+y^2)}{(1+x^2+y^2)^2}-\frac{2x}{1+y^2}&\frac{3x(1-y^2+x^2)}{(1+x^2+y^2)^2}-\frac{2y(1+x^2)}{(1+y^2)^2}\\2x&-2y\end{bmatrix}$$
Plugging in $(x,y)=(1,1)$, we get:
$$\begin{bmatrix}-\frac{2}{3}&\frac{4}{3}\\2&-2\end{bmatrix}$$
Now we need to find the eigenvalues of this matrix. We end up with the characteristic polynomial $\frac{3\lambda^2+8\lambda-4}{3}$. The roots of this polynomial are our eigenvalues, namely:
$$\frac{-4+2\sqrt{7}}{3}, \frac{-4-2\sqrt{7}}{3}$$
These are both real eigenvalues, one positive and one negative. We therefore have an (unstable) saddle point. You can verify this by graphing the system in pplane or some equivalent software.
