Let $f: \Bbb R^4 \to \Bbb R, f(x) =2x_1e^{x_2}+2x_2e^{x_3}+2x_3e^{x_4}$. Show that the level-set $S=f^{-1}(2)$ defines a regular surface.
From Wikipedia, I found this definition for regular surfaces.
Computing the gradient I have that $\nabla f(x)=(2e^{x_2}, 2x_1e^{x_2} + 2e^{x_3}, 2x_2e^{x_3}+ 2e^{x_4}, 2x_3e^{x_4})$. It's clear that $f$ is smooth. Now since $\nabla f \ne 0$ I get that by the by the implicit function theorem there exists an smooth function $g : U \to \Bbb R^3$ such that $\varphi(x_2,x_3,x_3) = (g(x_2,x_3,x_4), x_2,x_3,x_3)$.
Now according to the Wikipedia link, it seems that I would need to show that the rank of this parametrization $\varphi$ would be $3$. I'm a bit stuck here, how should I approach this?
We can avoid a messy computation in the application of the implicit function by using the inverse function theorem instead. Consider $f(x,y,z,w) =2xe^{y}+2ye^{z}+2ze^{w}-2.$
The level set we want to analyze is $S=\{(x,y,z,w):f(x,y,z,w)=0\}.$ Now,
$\partial_xf(x,y,z,w)=2e^y;\ \partial_yf(x,y,z,w)=2xe^y+2e^z$
$\partial_zf(x,y,z,w)=2ye^z+2e^w;\ \partial_wf(x,y,z,w)=2ze^w.$
There are no critical points, since $e^y\neq 0$ so every point $p$ on the level set is regular.
Now take the map $(x,y,z,w)\mapsto (f(x,y,z,w),y,z,w)$ and compute its Jacobian:
$\mathcal J=\begin{pmatrix} 2e^y &2xe^y+2e^z &2ye^z+2e^w &2ze^w \\ 0&1 &0 &0 \\ 0 &0 &1 &0 \\ 0 &0 &0 &1 \end{pmatrix}.$
It's easy to see that this is never zero so NOW we apply the inverse function theorem to get an open neighborhood $p\in U_p$ such that $(U_p,(f,y,z,w))$ is a chart in $\mathbb R^4$ with the property that on $U_p\cap S,$ the first coordinate vanishes. And this, by definition, makes $S$ a regular level set.