Imagine we have a function of position $x^i$ and velocity $v^i$ $f(x,v)$. Position and velocity are both functions of time $t$. If the function doesn't depend explicitely on time, then we have the identity
\begin{equation} \frac{d}{dt} f(x,v)= v^i \frac{\partial}{\partial x^i} f(x,v) \end{equation} Now imagine we define a new function
\begin{equation} G(x,v)=v^i \frac{\partial}{\partial x^i} f(x,v)=\frac{d}{dt} f(x,v) \end{equation}
and we want to derive $G$ respect to velocity. If we derive it using the first equation we get
\begin{equation} \frac{\partial}{\partial v^j}G(x,v)=\delta^{ij} \frac{\partial}{\partial x^i} f(x,v)+v^i\frac{\partial}{\partial v^j}\frac{\partial}{\partial x^i} f(x,v) \end{equation} % If we derive using the second equation instead we get
\begin{equation} \frac{\partial}{\partial v^j}G(x,v)=\frac{\partial}{\partial v^j}\frac{d}{dt} f(x,v) \end{equation} Now, I thought that these two derivatives commuted but that can't be right because then we would get
\begin{equation} \begin{aligned} \frac{\partial}{\partial v^j}G(x,v)&=\frac{\partial}{\partial v^j}\frac{d}{dt} f(x,v) \\ &=\frac{d}{dt}\frac{\partial}{\partial v^j}f(x,v) \\ &=v^i \frac{\partial}{\partial x^i }\frac{\partial}{\partial v^j} f(x,v) \end{aligned} \end{equation}
which is only the second term we obtained with the first identity. Does this mean that the commutator between $\frac{d}{dt}$ and $\frac{\partial}{\partial v^i}$ is
\begin{equation} [\frac{\partial}{\partial v^i},\frac{d}{dt}]=\frac{\partial}{\partial x^i} ? \end{equation}
