1. If $X$ is a vectorfield and $1$ is the constant function $x\to 1$, then $1^2=1$ and thus $X1=X1^2=2X1$, i.e. $X1=0$ and thus $Xc=0$ for any constant function $c$. 2. $X$ is a local operator, i.e. if $f$ and $g$ coincide in a neighborhood of $m$, then $Xf(m)=Xg(m)$: Indeed, put $h=f-g$ and let $\psi:M\rar[0,1]$ be smooth, $\psi(m)=1$ and $\psi|[f\neq g]=0$, then $h\psi=0$ and therefore $$ 0 =X(h\psi)(m) =h(m)X\psi(m)+\psi(m)Xh(m) =Xf(m)-Xg(m)~. $$ 3. By Taylor's Theorem we have for all $x\in B_r(m)\sbe M$ and $m_j=x_j(m)$: \begin{eqnarray*} f(x)-f(m) &=&\int_0^1\ttd tf(m+t(x-m))\,dt =\sum_j(x_j-m_j)h_j(x) \quad\mbox{where}\\ h_j(x)&=&\int_0^1\pa_jf(m+t(x-m))\,dt, \quad\mbox{i.e.}\quad h_j(m)=\pa_jf(m)~. \end{eqnarray*} Thus we have on $B_r(m)$: $f=f(m)+\sum (x_j-m_j)h_j$ and it follows by 1., 2. and the definition of a vector field that $$ Xf(m) =\sum Xh_j(m)(m_j-m_j)+h_j(m)Xx_j(m) =\sum_j Xx_j(m)\pa_jf(m) $$ i.e. $\z_j=Xx_j$. Conversely, if $X\colon=\sum\z_j\pa_j$, then by the product rule $X$ is obviously a vector field.