We have the following homework assignment, which result should be used for three other exercises. Unfortunately I didn't come up with the right idea. Maybe someone can help with hints, then I would elaborate within the forum. Thank you in advance! So here is the exercise:
Let $a$ be a regular value of $g$ and $M = g^{-1}(\{a\}) \neq \emptyset$ a $n-1$ dimensional submanifold of $\mathbb{R}^n$. Let $p \in M$ and $v \in T_pM$. Prove, that there exists a curve $\alpha: [-\epsilon, \epsilon] \rightarrow M$, such that $\alpha(0) = p$ and $\alpha'(0)=v$.
An outline (maybe not the easiest approach). You want a hint, so I let you fill in the details. Wlog let $a=0$
There is a curve $\gamma:(-\tau, \tau)\rightarrow \mathbb{R}^n$ such that $\gamma(0)= p$ and $\gamma^\prime(0)= v$. (You may take $\gamma(t) = p + tv$)
The problem is now, that the image of $\gamma$ may not be a subset of $M$. However, $g(\gamma(0))= g(p) = 0$ since $p\in M$.
Let $n := \frac{\nabla g}{||\nabla g||}$ and $\sigma(t,s) = \gamma(t) + sn(p)$. Consider the map $\xi:(s,t)\mapsto g(\sigma(t,s))$. It satisfies $\xi(0,0)=0 $ and
$$\frac{\partial}{\partial s}\xi |_{t,s=0}=\frac{\partial}{\partial s}g\circ \sigma |_{t,s=0}= dg(p)n(p)\neq 0$$
So the implicit function theorem implies that there is a (locally unique) map $s= h(t)$ such that $g(\sigma(t, h(t)))=0$
The curve you are after is $\sigma(t,h(t))$.
(Note: if you are not familiar with the topic this may look a bit like magic. It is not. You must know that $n$ is the unit normal to $M$ - what is happening here is that with $\sigma$ we disturb the curve $\gamma$ in direction towards $M$, so that we get a curve in $M$)