I have the following system:
$\dot{x}=y+x(x^4+2x^2y^2-4x^2+y^4-4y^2+4)(8-(x^2+y^2)^{\frac{3}{2}})^3$ $\dot{y}=-x+y(x^4+2x^2y^2-4x^2+y^4-4y^2+4)(8-(x^2+y^2)^{\frac{3}{2}})^3$
Using cylindrical coordinates this can be rewritten as:
$\dot{r}=r(r^2-2)^2(8-r^3)^3$
$\dot{\theta}=-1$
There is a unique fixed point at the origin and I would like to investigate its nature.
Using cartesian coordinates the jacobian matrix at the point (0,0) is $\begin{bmatrix}2048&&1\\-1&&2048\end{bmatrix}$ which suggests this point is an unstable focus (clockwise).
Using cylindrical coordinates the jacobian matrix at the point (0,0) is $\begin{bmatrix}2048&&0\\0&&0\end{bmatrix}$ which gives us no information.
So according to this information I would conclude that the fixed point (0,0) is an unstable focus. However when I plot the phase portrait of this system on mathematica it shows it as a star node. Is there a problem with my calculations? Is it a star node or unstable focus? Thank you.

The eigenvalues of your linearized system are $2048+i$ and $2048-i$. This suggests that the rate of growth in the radial direction will be much faster than the rate of rotation.
More generally, the general solution of the linear system
\begin{align} x' &= gx+y \\ y' &= -x+gy \end{align} is \begin{align} x(t) &= c_1 e^{g t} \cos (t)+c_2e^{g t} \sin(t) \\ y(t) &= c_2 e^{g t} \cos (t)-c_1e^{g t} \sin (t). \\ \end{align}
Thus, we can see that the larger $g$ is, the faster the rate of radial growth relative to the rotation. Here's an illustration of the phase portraits for various values of $g$: