Let me consider the map $$f:\Bbb{R}^2\rightarrow \Bbb{R}^3,~~~(x,y)\mapsto (x,y,xy)$$ I want to check if it is a regular surface patch.
We had the following definition:
$f:V\rightarrow \Bbb{R}^3$, $V\subset \Bbb{R}^2$ open is a regular surface patch if
$f$ is smooth, i.e. $f^{(n)}$ exists and is continuous for all $n\in \Bbb{N}$
$f$ is a homeomorphism on $f(V)$
the differential $df$ is injective
I mean clearly $f$ is smooth since $\frac{\partial}{\partial x} f(x,y)=(1,0,y)$, $\frac{\partial^{(n)}}{\partial x^{(n)}} f(x,y)=(0,0,0)$ for all $n\geq 2$ and $\frac{\partial}{\partial y} f(x,y)=(0,1,x)$, $\frac{\partial^{(n)}}{\partial y^{(n)}} f(x,y)=(0,0,0)$ for all $n\geq 2$ so all derivatives exists and are continuous. Furthermore we see that $$df=\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ y &x \end{pmatrix}$$ then if we consider the minor $$D:=\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$$ we see that $\det(D)=1\neq 0$ so $df$ is injective.
But now I have troubles in checking that it is a homeomorphism on $f(V)$. Can someone help me further?
Let $p : \mathbb R^3 \to \mathbb R^2, p(x,y,z) = (x,y)$. Then $p \circ f = id_{\mathbb R^2}$ which shows that $f$ is injective. Let $V' = f(V)$. Since $f$ is injective, we see that $f' : V \xrightarrow{f} V'$ is a continuous bijection. But $p(V') = p(f(V)) = V$, thus $p$ restricts to a continuous $p' : V' \to V$. By construction $p' \circ f' = id_V$, thus $(f')^{-1} = id_V \circ (f')^{-1} = p' \circ f' \circ (f')^{-1} = p'$ which shows that the inverse $(f')^{-1}$ of $f'$ is continuous.
Note that $V'$ is nothing else than the graph of the function $\phi : \mathbb R^2 \to \mathbb R, \phi(x,y) = xy$. It is a general theorem that the graph of a continuous function $u : X \to Y$ which is the subspace $$G(u) = \{ (x,u(x)) \mid x \in X \} \subset X \times Y$$ is homeomorphic to $X$ via $$h : X \to G(u), h(x) = (x,u(x)) .$$
You can easily generalize the above proof.
Also see https://math.stackexchange.com/q/4614228}.