Let $f \in \mathcal{C}^1(\mathbb{R}^3;\mathbb{R}^2)$ be a function such that $f(3,-1,2)=0$ and $Df(3,-1,2)=\begin{bmatrix} 1 & 2 & 1 \\ 1 & -1 & 1 \end{bmatrix}$. Show that there exists an open set $U \subseteq \mathbb{R}$ containing 3 and a function $g \in \mathcal{C}^1$ such that $f(x,g_1(x),g_2(x))=0$ for all $x \in U$, where $g(x)=(g_1(x),g_2(x))$.
Implicit function theorem: Let $U \subseteq \mathbb{R}^{k+n}$ be open, and let $f \in \mathcal{C}^r(U;\mathbb{R}^n)$. Write $f$ as $f(x,y)$, where $x \in \mathbb{R}^n$ and $ y \in \mathbb{R}^n$. Suppose $f(a,b)=0$, and suppose $\frac{\partial{f}}{\partial{y}}(a,b)$ is an invertible matrix. Then there exists a neighborhood $N \subseteq \mathbb{R}^k$ of $a$ and a function $g: N \rightarrow \mathbb{R}^n$ such that $g(a)=b$ and $f(x,g(x))=0$ for all $ x \in N$. Moreover, g is a $\mathcal{C}^r$ function.
With the implicit theorem in hand and the conditions given I don't think I have to do anything extra to explain right? Only state the theorem and say the conditions are satisfied, and then pretty much done?