I've a little problem
Suppose we have $$f,g:U \subset \mathbb{R}^n \rightarrow \mathbb{R} \in C^1,$$ where $g$ define $$\Lambda=\{x \in U: g(x)=0\}.$$
When I have to show that $ \nabla f(x_0) \space|| \space \nabla g(x_0)$ (for the regular point $x_0$) i have to consider some surface $r(p) \in \Lambda$ for which $\nabla f(x_0), \nabla g(x_0) \perp \frac{d}{dp}r(p_0)$ where $r(p_0)=x_0$, what if $U \subset \mathbb{R^3}$ and for every $r:V \subset \mathbb{R}^2 \rightarrow \Lambda \subset \mathbb{R}^3$ we get ${\rm rank}(\frac{d}{dp}r(p_0) ) \leq 1$?
This means that the space orthogonal to Jacobian matrix column of r have $\dim \gt 1$ and i can't conclude that $\nabla f$ and $\nabla g$ are parallels.
I'll do all steps: suppose we have the regular point $x_0$, then I can find a surface described by $r:V \subset \mathbb{R}^{n-1} \rightarrow \Lambda$, $V$ is a neighbor of $p_0$ and $r(p_0)=x_0$, for which $g\circ r(p)=0, \forall p \in V$. Now computing the derivative (i'll note $J_r$ as Jacobian of $r$): $$0=\frac{d}{dp}g\circ r (p)\space \big|_{p=p_0} =\nabla g(x_0)*J_{r}(p_0)$$
this is true also for the function $f$, indeed assuming that $x_0$ is a critical point for $f$ over the constrains $$0=\frac{d}{dp}f\circ r(p)\space \big|_{p=p_0}=\nabla f(x_0) * J_r (p_0) $$
now if $\mathrm{rank}(J_r(p_0))=n-1$ we have that the space orthogonal to the $J_r$ columns have $\dim =1$ then we can conclude that $\nabla f(x_0) \space ||\space \nabla g(x_0) $.
the question is: $\mathrm{rank}(J_r(p_0))=n-1$ always?
it's clear that for $f,g:\mathbb{R}^2\to\mathbb{R}$ the quantity $J_r$ is a vector and everything will be fine, what if there is $\mathbb{R}^n$, $n \gt 2$?