There exists $G$ s.t. $f \circ G$ is constant(proof completing)

Question

Problem: assume that $f\in C^1(\mathbb{R}^2,\mathbb{R})$, prove that there exists a continuous injection $G:\mathbb{R} \rightarrow \mathbb{R}^2$ s.t. $f \circ G$ is a constant function. I 've come up with an uncomleted idea, and I'd like you help to me complete it. And I am aware of this post being similar to mine, but notice that it doesn't requires $G$ to be defined on $\mathbb{R}$.
I think we could apply the implicit function theorem. When $\partial_x f\equiv \partial_y f \equiv0$ it is trivial. So I considered the special case when there exists $(a,b)$ s.t. $\frac{\partial f}{\partial y}(a,b) \neq0$, and in this restriction I can find such a $G$ in a small neighborhood. My attempt is as follows.
Let $(a,b)\in \mathbb{R}^2, c=f(a,b).$ Consider $F: \mathbb{R}^2 \rightarrow \mathbb{R},(x,y) \mapsto f(x,y)-c$. Then $F(a,b)=0,\frac{\partial F}{\partial y}(a,b) \neq0$, by the implicit function theorem there exist $\epsilon >0$ and a unique $\phi \in C^1(B_\epsilon(a),\mathbb{R})$ s.t. $\forall x \in B_\epsilon(a),F(x, \phi(x))=0,$ hence $f(x,\phi(x)) =c$. So we consider $G:B_\epsilon(a) \rightarrow \mathbb{R}^2,x \mapsto (x,\phi(x)).$ Then $\forall x\in B_\epsilon(a) ,f \circ G (x)=c$. And $G$ is clearly continuous and injective.
So my main question is: how could I "extend" the domain of $G$, i.e, $B_{\epsilon}(a)$ to $\mathbb{R}$?