Lie bracket of canonical vectors on tangent space to a point on a manifold is zero. Let M be a manifold and $T_p(M)$ be the tangent space at $p$, and $\phi$ a local chart around $p$. 
Let $$\left.\frac{\partial}{\partial\phi^1}\right|_{_p},\ \cdots\ ,\left.\frac{\partial}{\partial\phi^n}\right|_{_p}$$ be a basis of this vector space. Apparently the lie bracket of these basis vector is $0$, i.e. for instance, $$\left[\frac{\partial}{\partial\phi^1},\frac{\partial}{\partial\phi^2}\right]_pf=0, \quad\forall p, \forall f$$ smooth, i.e.
$$\frac{\partial}{\partial x^1}\left(\phi^{-1} \circ \frac{\partial f}{\partial \phi^2}\right)(\phi(p))=\frac{\partial}{\partial x^2}\left(\phi^{-1} \circ \frac{\partial f}{\partial \phi^1}\right)(\phi(p)).$$ 
This seems an easy exercise but I am not sure how to prove it properly...
Note that here $$\frac{\partial f}{\partial \phi^2} :M \rightarrow \mathbb{R},\quad p \mapsto \left.\frac{\partial}{\partial \phi^2}\right|_{p}(f) $$
 A: By definition of the Lie bracket
$$\left[\frac{\partial}{\partial\phi^i},\frac{\partial}{\partial\phi^j}\right]_p(f)=\frac{\partial}{\partial\phi^i} |_p(\frac{\partial}{\partial\phi^j} |_{(\cdot)}f)-\frac{\partial}{\partial\phi^j} |_p(\frac{\partial}{\partial\phi^i} |_{(\cdot)}f)$$
where $\frac{\partial}{\partial\phi^k} |_{(\cdot)}f$ denotes the smooth function $q\mapsto \frac{\partial}{\partial\phi^k} |_{q}f$.
Now since $ \frac{\partial}{\partial\phi^k}|_{(\cdot)}f=\frac{\partial}{\partial x^k}|_{\phi(\cdot)}(f\circ\phi^{-1})$
$$
\frac{\partial}{\partial\phi^i} |_p(\frac{\partial}{\partial\phi^j} |_{(\cdot)}f)
=\frac{\partial}{\partial x^i} |_{\phi (p)}((\frac{\partial}{\partial\phi^j} |_{(\cdot)}f)\circ\phi^{-1})
\\=\frac{\partial}{\partial x^i} |_{\phi(p)}(\frac{\partial}{\partial\phi^j} |_{\phi^{-1}(\cdot)}f)
=\frac{\partial}{\partial x^i} |_{\phi(p)}(\frac{\partial}{\partial x^j} |_{(\cdot)}(f\circ\phi^{-1}))
=\frac{\partial^2}{\partial x^i\partial x^j}(f\circ\phi^{-1})_{|\phi(p)}$$
Similar
$$
\frac{\partial}{\partial\phi^j} |_p(\frac{\partial}{\partial\phi^i} |_{(\cdot)}f)
=\frac{\partial^2}{\partial x^j\partial x^i}(f\circ\phi^{-1})_{|\phi(p)}
$$
so by Schwarz's theorem we have  $\left[\frac{\partial}{\partial\phi^i},\frac{\partial}{\partial\phi^j}\right]_p(f)=0$.
