Finding the vector field of ODE in cartesian coordinate 
The way I approached the problem is first by solving a system of equations:
$$
r_x \frac{dx}{dt} + r_y\frac{dy}{dt} = \frac{dr}{dt} = -r
$$
$$
\theta_x \frac{dx}{dt} + \theta_y\frac{dy}{dt} = \frac{2}{\ln(x^2+y^2)}
$$
where $r = \sqrt{x^2+y^2}$ and $\theta = tan^{-1}\frac yx$. The problem is after solving for $\frac {dx} {dt}$ and $\frac{dy}{dt}$ I got the following vector field:
$$
\left(\frac{dx}{dt}, \frac{dy}{dt}\right) = \left(-x-\frac{2y}{ln(x^2+y^2)},-y+\frac{2x}{ln(x^2+y^2)}\right)
$$
And this mess doesn't seem to be differentiable at $(0,0)$. Anyone got some alternative approach or can point out where I made a mistake? Thanks!
EDIT: I found somewhere I made a mistake in getting to the vector field so here I edited the problem to show the new (hopefully correct) vector field obtained. However it is still not differentiable at the origin.
 A: Hint: You can solve the system
$$\dot{r}=-r \implies r(t) = r_0e^{-t}$$
$$\dot{\theta} = \dfrac{1}{\ln r}=\dfrac{1}{\ln r_0 - t} \implies \theta(t) = -\ln\left[\ln r_0 - t \right] + c$$
We must assume $t\neq \ln r_0$. Then the cartesian coordinates representation will be
$$x(t) = r(t)\cos \theta(t)$$
$$y(t) = r(t) \sin \theta(t).$$
A: You could also explicitly go for $\dot x$ and $\dot y$, using
\begin{align}
\dot x &= \frac{d}{dt}(r\cos\theta) = \dot r\cos\theta - r\dot\theta \sin\theta\\
\dot y &= \frac{d}{dt}(r\sin\theta) = \dot r\sin\theta + r\dot\theta \cos\theta
\end{align}
Since $ \cos\theta = \frac{x}{r} $ and $ \sin\theta = \frac{y}{r} $,
this is pretty straightforward.
Your answer seems correct; it is undefined at the origin because of the $\ln r$
term in $\dot\theta$. But note that a continuous extension is mentioned:
$ f(0, 0) = (0, 0) $ by definition.
Since $ \ln r \to -\infty $ as $ r \to 0 $, $ \frac{1}{\ln r} \to 0 $ when
approaching the origin, so $f$ is indeed continuous (and differentiable) at the origin.
