If $f:\mathbb{R}^2\to\mathbb{R}^1$ is of class $C^1$, show that $f$ is not one-to-one. If $f:\mathbb{R}^2\to\mathbb{R}^1$ is of class $C^1$, show that $f$ is not one-to-one. [Hint: If $Df(x) = 0$ for all $x$, then $f$ is constant. If $Df(x_0)\neq0$, apply the implicit function theorem.] 
Clearly there are two cases, if $Df(x)=0$ for all $x\in \mathbb{R}^2$ then $f$ is constant and therefore can not be one to one.
If there is a $x_0\in\mathbb{R}^2$ such that $Df(x_0)\neq 0$ then how can I use the implicit function theorem knowing that in order to apply it I have to ensure that $f(x_0)=0$ but this is not given in the problem? Here is the version of the theorem of the implicit function that I am using, thank you very much.

 A: Suppose that $Df(x_0,y_0)\neq 0$. You can suppose without restricting the generality that ${{\partial f}\over{\partial y}}\neq 0$, let $h(x,y)=f(x,y)-f(x_0,y_0), {{\partial h}\over{\partial y}}={{\partial f}\over{\partial y}}\neq 0$, the implicit function theorem implies that  there exists a neighborhood $I$ of $y_0$, a function $h:I\rightarrow \mathbb{R}^2$ such that $h(x,g(x))=f(x,g(x))-f(x_0,y_0)=0$.
A: It suffices to prove the theorem in the case if $f(x_0) = 0$, because if you have an arbitrary injective $C^1$ map $f$ and $x_0$ where the $Df(x) \neq 0$, then you can apply the special case to $f(x)-f(x_0)$, which does satisfy the conditions for the implicit function theorem.
A: Since the OP asks in comments what happens if $f$ is just differentiable, this may be useful:
The result holds even if $f$ is just continuous, from the invariance of domain theorem. Indeed, suppose $f$ is one-to-one. By invariance of domain, $\iota \circ f: \mathbb{R}^2 \to \mathbb{R}^2$ is open, where $\iota:\mathbb{R} \to \mathbb{R}^2$ is the standard inclusion. However, since the image is contained in $\mathbb{R} \times \{0\}$, the map $\iota \circ f$ can't be open.
