$\newcommand{\Reals}{\mathbf{R}}$So here is an exercise from classical differential geometry;
Prove that if $h_1:(a,b) \to \Reals$ and $h_2:(a,b) \to \Reals$ are $C^2$ functions in a non-zero interval $(a,b) \subset \Reals$, then the parametric equation $r:(a,b) \to \Reals^3$ defined by $r(x) =xi + h_1(x)j + h_2(x)k$ is regular. Explain why every plane $x = c$, with $c \in (a,b)$, intersects with this curve in a single point.
So this is the first part of the exercise. To prove that it's a regular curve, I check the vector of speed which is $$ \frac{dr}{dx} = [1, h_1'(x), h_2'(x)] \neq [0,0,0]. $$ So it's a regular curve.
For the other part is this right? To find their intersection I replace the parameters of the curve in the equation of the plane. So I have that their intersection is $x = c$ and that means that their intersection is the single point $(c,0,0)$ and thus every plane $x = c$ intersects with the curve in a single point. Is that false?
And now the part that I have absolutely no idea what I have to do;
Prove the opposite that if every plane $x = c$ stable intersects with a regular curve in maximum one point transversely, then we can have functions $h_1$, $h_2$ like before, so that the curve can be parameterized through the $x$-axis.
Any help would be appreciated.
(i) Write your curve $\gamma$ in the form $$\gamma:\quad t\mapsto{\bf h}(t):=\bigl(t,h_2(t),h_3(t)\bigr)\qquad(a<t<b)\ .\tag{1}$$ Then ${\bf h}'(t)=\bigl(1,h_2'(t),h_3'(t)\bigr)\ne{\bf 0}$ for all $t$, hence the representation is regular.
(ii) Intersecting $\gamma$ with the plane $x=c$ enforces $t=c$. Therefore we have exactly one point of intersection, namely the point ${\bf r}_c:=\bigl(c, h_2(c),h_3(c)\bigr)$. (You had this wrong.)
(iii) Now let $$\gamma:\quad t\mapsto{\bf h}(t):=\bigl(h_1(t),h_2(t),h_3(t)\bigr)\qquad(a<t<b)\tag{2}$$ be a regular $C^1$-curve, and consider a $t_0\in(a,b)$. It is assumed that $\gamma$ intersects the plane $\pi: \ x=h_1(t_0)$ transversally in the point ${\bf p}={\bf h}(t_0)=\bigl(h_1(t_0),h_2(t_0),h_3(t_0)\bigr)$. "Transversal" means that the two vectors ${\bf e}_2=(0,1,0)$ and ${\bf e}_3=(0,0,1)$ spanning $\pi$, together with ${\bf h}'(t_0)$, span the full three-dimensional tangent space $T_{\bf p}$. It follows that necessarily $h_1'(t_0)\ne 0$. Since $h_1'$ is assumed continuous this implies that $h_1'$ has constant sign on $(a,b)$, hence we may assume that $h_1$ is, e.g., strictly increasing. Moreover the function $h_1$ has a $C^1$-inverse $g: \ \tau\mapsto t=g(\tau)$, defined in $U:=h_1\bigl((a,b)\bigr)$. We therefore can replace the parametrization $(2)$ of $\gamma$ by $$\gamma:\quad \tau \mapsto\hat {\bf h}(\tau)={\bf h}\bigl(g(\tau)\bigr):=\bigl(\tau,h_2(g(\tau)),h_3(g(\tau))\bigr)\qquad(\tau\in U)\ ,$$ which is of the form $(1)$.