Stationary points and linearisation of non-linear system So, the problem is:

Find and discuss the behavior of the stationary points of the system :
$$ x'=-y+x\cdot (x^2+y^2)\cdot \sin\sqrt{x^2+y^2} =f(x,y)$$ 
$$ y'=x+y\cdot (x^2+y^2)\cdot \sin\sqrt{x^2+y^2}=g(x,y)$$

So in the beggining I Linearised the non-linear system, using the limits:
$$ \lim_{ r\to 0}\frac{f_1(x,y)}{r}=0 $$ and $$\lim_{r \to 0}\frac{g_1(x,y)}{r}=0$$ 
where $$f_1(x,y)=x\cdot (x^2+y^2)\cdot \sin\sqrt{x^2+y^2}$$ and 
$$g_1(x,y)=y\cdot (x^2+y^2)\cdot \sin\sqrt{x^2+y^2}$$
I linearise the non-linear system and reach it to the form:
$$x'=-y$$ and $$y'=x$$
but the problem is that i don't know how to calculate the stationary points from this non-linear system so to proceed with the next question of my problem .I also know that when i find the stationary points i have to take the jacobian so to characterise my stationary points.I would really, really appreciate any thorough help or hints/tips.
Thanks in advance!
 A: In order to get the critical points, we need to solve the system : 
$$\begin{cases} -y+x\cdot (x^2+y^2)\cdot \sin\sqrt{x^2+y^2} =0 \\ \space \space\space x+y\cdot (x^2+y^2)\cdot \sin\sqrt{x^2+y^2}=0\end{cases}$$
The obvious solution is $x=0,y=0$, which means that $O(0,0)$ is a Stanionary Point of your system.
To prove that it's unique, let's assume $x,y\neq 0$ and multiply the first equation by $-y$ and the second one by $x$. We then get :
$$\begin{cases} y^2 -xy(x^2+y^2)\sin(\sqrt{x^2+y^2}) = 0 \\ x^2 + xy(x^2 + y^2)\sin(\sqrt{x^2 + y^2}) = 0\end{cases}$$
By adding these two, we obviously get : 
$$ x^2 + y^2 = 0 $$
which can only hold for $x=y=0$, which is not compatible with our initial hypothesis.
Now, let's assume $x\neq 0,y=0$. We then get (through the initial given system) :
$$\begin{cases} x^3\cdot \sin\sqrt{x^2}=0\\x=0\end{cases}$$
which is also not compatible, since we had let $x\neq 0$.
Letting $x=0,y\neq 0$ gets us to :
$$\begin{cases} y=0 \\ y^3\sin\sqrt{y^2}=0\end{cases}$$
which is as well not compatible, since we had let $y\neq0$.
Thus, the only solution is : $\begin{cases} x=0 \\ y=0 \end{cases}$ and your only stationary point is the $O(0,0)$.
