Finding regular points and values of $f$ 
let $f\in C^1(\mathbb{R}^3, \mathbb{R}^2)$, $f(x,y,z)=(x^2+y^2,
 y^2+z^2)$. Find the set of regular points and the set of the regular
  values.

My idea:
Computing the Jacobian, $J_f(x,y,z)=2\begin{pmatrix} x & y & 0 \\ 0 & y & z \end{pmatrix}$. The Jacobian has full rank iff $x\neq 0$ or $z\neq 0$. So, the set of regular points is $R_f:=\{(x,y,z)\in \mathbb{R}^3 : x,z\in \mathbb{R}\backslash \{0\}\, \land \, y\in \mathbb{R}\}$. Is that correct and how can I find the set of the regular values?
 A: First, a tiny tweak on your condition for regular points. You said that iff $x \neq 0$ or $z \neq 0$, the Jacobian has full rank. But as Maximilian Janisch pointed out, this is not true! In fact, we must have that at least two of $x$, $y$, and $z$ are not zero. (Fortunately, this mistake doesn't change the result about regular values).
Note that the range of $f$ is a subset of ordered pairs $(u, v)$ with $u \geq 0$ and $v \geq 0$. I claim that every point except $(0,0)$ is regular. Why?
Suppose we have some arbitrary $(u, v) \neq (0,0)$. We want to show that there exists some $(x, y, z)$ with $x \neq 0$ or $z \neq 0$ that maps to it. We consider two cases:


*

*If $u = v$, then we can just choose $x = z = \sqrt{u} = \sqrt{v}$ and $y = 0$. Because $u, v \neq 0$, we know $x, z \neq 0$, and thus we are assured that this is a regular value.

*If $u \neq v$, then WLOG assume $u > v$. Choose $y = \sqrt{v}$, $z = 0$, and then choose $x = \sqrt{u^2 - v^2}$, which is again well-defined, as $u > v$ by assumption so $x > 0$ and hence both $y$ and $x$ are strictly positive, and $(x,y,z)$ is a regular point.


Of course, $(u, v) = (0,0)$ implies $x = y = z = 0$, as $(x, y, z) \in \mathbb{R}^3$, so it cannot be regular.
A: Since $\begin{pmatrix} x & y & 0 \\ 0 & y & z \end{pmatrix}\sim \begin{pmatrix} x & 0 & 0 \\ -x & 0 & -z \end{pmatrix}\sim \begin{pmatrix} x & 0 & 0 \\ 0 & 0 & -z \end{pmatrix}$, if either $x=0$ or $z=0$ then $(x,y,z)$ is not a regular point so the set of regular points is $\{(x,y,z):x,z\neq 0\}$.
The regular values are a subset of the open first quadrant $Q_1$ of $\mathbb R^2$. To see that they fill this set, it is enough to note that $f(x,0,z)=(x^2,z^2)$ is onto $Q_1.$
