Show that no application $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, of $C^k$ class, $k \geq 1$ can be injective How can I proof this:

Show that no application $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, of $C^k$ class, $k \geq 1$ can be injective, i.e., there are $A,B \in \mathbb{R}^2$ such that $A \neq B$ and $f(A) = f(B)$ (Hint: show that given $(a,b) \in \mathbb{R}^2$ in every open neighbourhood of $(a,b)$ there are points $(c,d),(c_1,d_1)$ where ${\partial}_1f(c,d) \neq 0$ or ${\partial}_1f(c_1,d_1) \neq 0$).

The problem is that I have no idea how to use the hint.
Thanks for all your comments.

If $df =0$ everywhere, then $f(a, b)$ is constant and hence cannot be injective.  On the other hand, if at some point $(a_0, b_0) \in \Bbb R$ we have $df(a_0,  b_0) \ne 0$, we must also have
$f_a(a_0, b_0) = \dfrac{\partial f}{\partial a}(a_0, b_0) \ne 0 \tag{1}$
or
$f_b(a_0, b_0) = \dfrac{\partial f}{\partial b}(a_0, b_0) \ne 0. \tag{2}$
If (1) holds, then by the implicit function theorem there are open intervals $I, J \subset \Bbb R$ with $(a_0, b_0) \in I \times J$ and a function $g:J \to I$ such that $f(g(b), b) = f(a_0, b_0)$ for $b \in J$, with the corresponding result in the case that (2) applies.  Thus (1) leads to the conclusion that there is a continuum of points $(g(b), b) \in \Bbb R^2$, $b \in J$, for which $f(g(b), b) = f(a_0, b_0)$; $f$ cannot be injective under such circumstances; the analogous result with $f(a, g(a)) = f(a_0, b_0)$, $a \in I$,  follows from (2).  Thus we see that no $f \in C^k(\Bbb R^2, \Bbb R)$, $k \ge 1$, can be injective.  QED.
P.S. I don't quite get the hints myself.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
