What information can one get from $f(x,y)\geq -3x+4y$ provided that $f$ is continuously differentiable near $(0,0)$? 
Let $V$ be a neighborhood of the origin in ${\Bbb R}^2$ and $f:V\to{\Bbb R}$ be continuously differentiable. Assume that $f(0,0)=0$ and $f(x,y)\geq -3x+4y$ for $(x,y)\in V$. Prove that there is a neighborhood $U$ of the origin in ${\Bbb R}^2$ and a positive number $\epsilon$ such that, if $(x_1,y_1),(x_2,y_2)\in U$ and $f(x_1,y_1)=f(x_2,y_2)=0$, then
  $$
|y_1-y_2|\geq\epsilon|x_1-x_2|.
$$

Using the assumption, we have
$$
f(x)=f'(0)x+o(\|x\|)
$$
which gives the local behavior of $f$ near the origin. But how the inequality $f(x,y)\geq -3x+4y$ would be used here?
 A: The inequality means that $f$ always lies above a certain plane determined by the value of the linear combination. Since $f$ is differentiable, it has a tangent plane at the origin, so if the two planes are not the same, you can find a suitable point close to the origin where $f$ is asymptotically close to the second plane and hence lies below the first plane, and hence a contradiction.
A: Note that $f_y(0,0) \geq 4,$ and hence $Df (0,0) \neq 0.$ Hence the implicit function theorem tells you that that there is a neighborhood $(-\epsilon,\epsilon)$ of $0 \in \mathbb{R}$ and a $C^1$ diffeomorphism $g: (-\epsilon, \epsilon) \to g (-\epsilon, \epsilon)$ such that $f(x,y) = 0 \Rightarrow y = g(x),$ for all appropriate $x,y.$ 
Hence given $(x_1,y_1)$ and $(x_2,y_2)$ with $x_1 \neq x_2$ and those points sufficiently close to $0,$ and satisfy $f(x_j,y_j) = 0, j = 1,2$ then we see $y_1 \neq y_2$ (by the local injectivity part of the implicit function theorem). 
Hence $y_1 - y_2 = g(x_1) - g(x_2) = g'(\xi) (x_1 - x_2),$ for some point $\xi \in (x_1,x_2).$ 
Then note that $g' \neq 0$ on $(-\epsilon,\epsilon).$ The conclusion follows since $g$ is $C^1.$ ($g'$ has a positive minimum on a compact subset of $(-\epsilon,\epsilon)$ containing $x_1,x_2.$ )
