Show that if $f : \mathbb{R}^{2} \to \mathbb{R}$ continuously differetiable then $f$ is not inyective Well my question this time is:
How to show that $f : \mathbb{R}^{2} \to \mathbb{R}$ continuously differetiable then $f$ is not inyective 
I was trying to consider the function $g(x,y)=(f(x,y),y)$, but I don't know how to proceed with $g$, I know I have to use the implicit function theorem but when I was thinking of it, I remember that nearly $(x_0,y_0)$, $f$ must be zero, well $f(x_0,y_0)=0$, then the only tool I thought it could help I can't apply it.
Can someone help me please with this problem?, and once you have it for this case, Can we generalize to $R^{m}$ and $R^{n}$?(of course $m<n$) 
Thanks a lot in advance
My implicit function theorem:
Let $A \subset R^{m} \times R^{n}$ open, $(x_0,y_0) \in A$, $f:A \to R^{m}$ continously differentiable nearly $(x_0,y_0)$ and  $f(x_0,y_0)=0$. Let M be the matrix of $m \times m$ given by:
$$D_{n+j}f^{i}(x_0,y_0)$$
and suppose that $\det(M) \not= 0$, then there exists an open set $U \subset R^{n}$ $x_0 \in U$, and an open $V \subset R^{m}$, $y_0 \in V $, such that, for each $x$ there exists a unique $g(x) \in V$ such that $f(x,g(x))=0$, $g$ differentiable
I don't know Topology, or more than introduction to analysis and a little bit of multivariable analysis, thanks a lot for your help 

Suppose that $f$ is injective. Then $f(0,0)\ne f(1,0)$.
On $[0,1]$, consider the functions $g(t)=f(t,t-t^2)$ and $h(t)=f(t,t^2-t)$. Note that $g(0)=h(0)=f(0,0)$ and $g(1)=h(1)=f(1,0)$
The Intermediate Value Theorem says that for some $t_g\in(0,1)$, $g(t_g)=\frac{f(0,0)+f(1,0)}2$ and for some $t_h\in(0,1)$, $h(t_h)=\frac{f(0,0)+f(1,0)}2$. This means that
$$
f(t_g,t_g-t_g^2)=g(t_g)=\frac{f(0,0)+f(1,0)}2=h(t_h)=f(t_h,t_h^2-t_h)
$$
However, $t_g-t_g^2\gt0\gt t_h^2-t_h$. Thus, we have $f(t_g,t_g-t_g^2)=f(t_h,t_h^2-t_h)$ yet $(t_g,t_g-t_g^2)\ne(t_h,t_h^2-t_h)$. Thus, $f$ is not injective.
