Lie Bracket of polar coordinate vector fields. I've been thinking about this problem for some time, yet I am unable to solve it or find a resource online. I know that $\frac{\partial}{\partial r}$ and $\frac{\partial}{\partial \theta}$ are coordinate vector fields and their bracket should be zero, but that reasoning doesn't seem rigorous enough. Here is the problem:
Let $\frac{\partial}{\partial r}$ be the normalized position field on $\mathbb R^2 \setminus \{ 0 \} $. That is, when viewed as a map $\mathbb R^2 \setminus \{ 0 \} \longrightarrow \mathbb R^2 \setminus \{ 0 \}$, $\frac{\partial}{\partial r}$ is
$$ \frac{\partial}{\partial r} (x,y) = \frac{(x,y)}{|(x,y)|}. $$
Let $\frac{\partial}{\partial \theta}$ be the velocity field of the $S^1$-action on $\mathbb R^2$ given by the counterclockwise rotation
$$ S^1\times\mathbb R^2 \longrightarrow \mathbb R^2 $$
where
$$ \left((\cos\theta,\sin\theta),(x,y)\right) \longmapsto 
\begin{pmatrix} \cos\theta & -\sin\theta \\\ \sin\theta & \cos\theta \end{pmatrix}
\begin{pmatrix}
x \\\ y
\end{pmatrix}.$$
That is,
$$ \frac{\partial}{\partial \theta}(x,y) = \frac{\partial}{\partial \theta} \begin{pmatrix} \cos\theta & -\sin\theta \\\ \sin\theta & \cos\theta \end{pmatrix} \left.\begin{pmatrix}
x \\\ y
\end{pmatrix} \right|_{\theta=0}. $$
Show that $ [\frac{\partial}{\partial \theta},\frac{\partial}{\partial r}] = 0 $.

$$[{\partial _\theta},{\partial_r}](x,y)={\partial_ \theta}\left({\partial_r}(x,y)\right)-\partial_r\left(\partial_\theta(x,y)\right)$$
$$=\partial_\theta{(x,y)\over|(x,y)|}-\partial_r\left(-x\sin(\theta)-y\cos(\theta), x\cos(\theta)-y\sin(\theta)\right)$$
$$={1\over |(x,y)|}\left(-x\sin(\theta)-y\cos(\theta), x\cos(\theta)-y\sin(\theta)\right)-{1 \over \scriptsize|\left(-x\sin(\theta)-y\cos(\theta), x\cos(\theta)-y\sin(\theta)\right)|}\left(-x\sin(\theta)-y\cos(\theta), x\cos(\theta)-y\sin(\theta)\right) $$
$$={1\over |(x,y)|}\left(x\sin(\theta)-y\cos(\theta), x\cos(\theta)-y\sin(\theta)\right)-{1 \over |(x,y)|}\left(-x\sin(\theta)-y\cos(\theta), x\cos(\theta)-y\sin(\theta)\right) $$
$$=0$$
