Show that at a zero $x$, the derivative $dv_x:T_x(X)\rightarrow \mathbb{R}^N$ actually carries $T_x(X)$ into itself.

A vector field $v$ on a manifold $X$ in $\mathbb{R}^N$ is a particular type of map $v:X \rightarrow \mathbb{R}^N$. Show that at a zero $x$, the derivative $dv_x:T_x(X)\rightarrow \mathbb{R}^N$ actually carries $T_x(X)$ into itself.

There is similar post in MSE:

Show that at a zero $x$, the derivative $d(\vec v_x ) :T_x (X)→R^N$ actually carries $T_x (X)$ into itself.

But its not helpful at all because the comments say that the problem or hint is wrong as I understand.

First of all I am confused by the term "at a zero $x$". What does that mean? Moreover honestly I have no idea how to approach this problem. Any help is appreciated!

For $X$ $n$-dimensional submanifold of $\mathbb{R}^N$ you can show it this way. Let $(U,\phi)$ be a local chart centered at $x$, where $U\in X$, $\phi:\mathbb{R}^n\rightarrow X$. Call $z=(z_1,\dots,z_n)$ the coordinates given by $\phi$, let $\bar{z}\in\mathbb{R}^n$ be such that $\phi(\bar{z})=x$ and call $\{\frac{\partial\phi}{\partial z^1}\rvert_{\phi(z)},\dots,\frac{\partial\phi}{\partial z^n}\rvert_{\phi(z)}\}$ the basis of the tangent space induced by this chart. This can be thought as a subspace of $\mathbb{R}^N$ of dimension $n$. Thus in coordinates we can write
$$v(\phi(z))=\sum_{j=1}^n v^j(z)\frac{\partial\phi}{\partial z^j}\rvert_{\phi(z)},$$ where $v^j\in C^{\infty}(\mathbb{R}^n)$ (because $v(y)$ belongs to the tangent space of $X$ at $y$ for every $y\in X$). To conclude it's enough to show that $dv_{\phi(\bar{z})}\left(\frac{\partial\phi}{\partial z^j}\rvert_{\phi(\bar{z})}\right)$ can be written as a linear combination of $\frac{\partial\phi}{\partial z^1}\rvert_{\phi(\bar{z})},\dots,\frac{\partial\phi}{\partial z^n}\rvert_{\phi(\bar{z})}$. $$dv_{\phi(\bar{z})}\left(\frac{\partial\phi}{\partial z^j}\rvert_{\phi(\bar{z})}\right)=\left(\frac{\partial (v\circ\phi)(z)}{\partial z^j}\right)\rvert_{\bar{z}}=\frac{\partial}{\partial z^j}\left(\sum_{i=1}^n v^i(z)\frac{\partial\phi}{\partial z^i}\rvert_{\phi(z)}\right)\rvert_{\bar{z}}=\left(\sum_{i=1}^n\frac{\partial v^i}{\partial z^j}(\bar{z})\frac{\partial\phi}{\partial z^i}\rvert_{\phi(\bar{z})}\right)+\left(\sum_{i=1}^n v^i(\bar{z}) \frac{\partial}{\partial z^j}\frac{\partial\phi}{\partial z^i}\rvert_{\phi(\bar{z})}\right)=\left(\sum_{i=1}^n\frac{\partial v^i}{\partial z^j}(\bar{z})\frac{\partial\phi}{\partial z^i}\rvert_{\phi(\bar{z})}\right).$$ There is a subtle indetification of the tangent spaces of $\mathbb{R}^N$ here. Indeed the above is correct, but on the other hand, $$dv_{\phi(\bar{z})}\left(\frac{\partial\phi}{\partial z^j}\rvert_{\phi(\bar{z})}\right)=\sum_{i=1}^N a_j^ie_i,$$ where $\{e_1,\dots,e_N\}$ is the basis of $T_0\mathbb{R}^N$. But since the tangent spaces of $\mathbb{R}^N$ are canonically isomorphic this is not a problem.