Show that preimage is an embedded surface I have given a function
$$f:\mathbb{R}^4 \to \mathbb{R}^3: (x,y,z,u) \mapsto (xz-y^2, yu-z^2,xu-yz)$$
and I want to show that $f^{-1}(0)\setminus\{0\}$ is an embedded surface.
If it would be an embedded curve I could verify that $f$ is an submersion to get a 1-dim submanifold. But I don't know how to show that this is an embedded surface.

I saw this problem from Amann's Analysis II, page 257.
Note that
$$\begin{aligned}
y^2&=xz\\
z^2&=yu\\
yz&=xu.
\end{aligned}$$
If $yz\neq 0$, then
$$\begin{aligned}
y^2\cdot yz=xz\cdot xu&\implies y=x^{2/3}\cdot u^{1/3}\\
z^2\cdot yz=yu\cdot xu&\implies z=x^{1/3}\cdot u^{2/3}.
\end{aligned}$$
In the case $yz=0$, then it is easy to see that $xu=0$ and $y=0=z$, where the above relations also hold.
Let $(a,b)=(x^{1/3},u^{1/3})$ and $$g:(a,b)\mapsto (a^3,a^2b,ab^2,b^3).$$
It is clear that $g$ is a homeomorphism.
And $$\partial g=\begin{pmatrix} 3a^2& 0\\ 2ab&a^2\\ b^2&2ab\\ 0&3b^2 \end{pmatrix}$$ and $g$ is an embedding when $(a,b)\neq (0,0)$.
It follows that $g(\mathbb R^2\setminus \{(0,0)\})=f^{-1}(0)\setminus \{0\}$ is a embedded curve.
