Let $F:\mathbb{R}^3\to \mathbb{R}$ be a $C^1$-function and suppose that $(dF)(x,y,z)\not=0$ wherever $F(x,y,z)=0$. Call $$O = \{(x,y,z)\mid F(x,y,z)=0\}.$$ Then, for every point $p\in O$, there is an open neighborhood $V\subseteq \mathbb{R}^3$ of $p$ and an open neighborhood $U\subseteq\mathbb{R}^3$ of $(0,0,0)$ and a bijective $C^1$-map $\varphi:U\to V$ with a $C^1$-inverse such that $\varphi(U\cap(\mathbb{R}^2\times\{0\}))=V\cap O.$
Proof:
Choose an arbitrary point $p\in O$. We first note that the rank of $(dF)(p)$ is $1$, since $(dF)(p)$ is a $1\times 3$ non-zero matrix. Hence, since it only has one row, it has full rank i.e it has rank $1$.
Now it is possible to parameterize $O$ with a $C^1$-map $$g:U_0\subseteq \mathbb{R}^2\to \mathbb{R}^3$$ such that $g(0,0) = p$, where $U_0\subseteq \mathbb{R}^2$ is an open neighborhood of $(0,0)$ and $V_0\subseteq \mathbb{R}^3$ is an open neighborhood of $p$ and $g(U_0)=V_0\cap O.$ Since $F$ is $C^1$, so too is $g$.
We now define the map $$\tilde{\varphi}:U_0\times\mathbb{R}\to\mathbb{R}^3:(x,y,z)\mapsto g(x,y)+(z,z,z).$$ It is easily verified that for all $(x,y,z)\in U_0\times \mathbb{R}$, the total derivative $(d\tilde{\varphi})(x,y,z)$ is invertible and that $\tilde{\varphi}$ is a $C^1$-map.
The inverse function theorem now tells us that there are open neighborhoods $U\subseteq U_0\times\mathbb{R}$ of $(0,0,0)$ and $V\subseteq \mathbb{R}^3$ of $p=\tilde{\varphi}(0,0,0)$ such that $\tilde{\varphi}\mid_U:U\to V$ is $C^1$-bijection with a $C^1$-inverse. Call this restriction $\varphi$. Then it is easily seen that $\varphi(U\cap(\mathbb{R}^2\times\{0\}))=V\cap O$.
Could someone verify if this proof is correct?