Show that at a zero $x$, the derivative $d(\vec v_x ) :T_x (X)→R^N$ actually carries $T_x (X)$ into itself.
I know that we have the vector field $\vec v:X\to R^n$. If $X=R^n \times \{0\}$, then the claim is true obviously..
The book tells me to find a local parametrization, so let $\phi:U\to R^n$ where $U$ is the open neighborhood in $X$. But I can't see how this gonna help me