How do I prove that there is a neighbourhood $U$ of the orign in $\mathbb{R}^2$ and, $|y_{2}-y_{1}|\geq \epsilon|x_{2}-x_{1}|$. Let V be a neighborhood of the origin in $\mathbb{R}^2$, and $f: V \rightarrow \mathbb{R}$ 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 $\mathbb{R}^2$ 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 $|y_{2}-y_{1}|\geq \epsilon|x_{2}-x_{1}|$.
This is my idea to solve this question:
Since f is a continuously differentiable, and $f(0,0)=0$.
and $f_{y}\geq4$ at $(0,0)$. By implicit function theorem, there exists a neighborhood $W$ of $0$, and a neighborhood $U$ of $0$, and there a unique continuously differentiable map $g:W \rightarrow U$, such that $g(0)=0$, and $f(x,g(x))=0$ for $x \in W$.
for second part I am thinking to use Mean value theorem, I stuck at $|g(x_{2})-g(x_{1})|=|g^{'}(c)||x_{2}-x_{1}|$.
Can anyone please suggest me the direction of this solution?
Is my answer on a right way?
Well, the IFT also tells you how much $g'(0)$ is. Since $g'(0) \neq 0$ and $g$ is continuous at $0$, there is an $\epsilon &gt;0$ such that $g'(c) \geq \epsilon$ for all $c$ in an open interval containing $0$. Also add to this the fact that $(x,g(x))$ are ALL of the zeros of the function $f$ in a neighborhood, that is, if $f(x,y)=0$, then $y=g(x)$.
You have a good idea. Just prove that $g'(0)\ne 0.$ Then $|g'| &gt;a$ for some positve $a$ in a neighborhood of $0,$ and that implies what you want.
To show $g'(0)\ne 0,$ note that $f(x,g(x)) =  0$ for $x$ near $0.$ Differentiate this at $x=0$ to see
$$\tag 1 D_x f(0,0)\cdot 1 + D_y f(0,0)\cdot g'(0)=0.$$
Now because $f(0,0)=0$ and $f(x,y)\ge -3x+4y$ you can see that for $x&lt;0,$ $f(x,0)\ge -3x = 3|x|.$ This implies $|D_x f(0,0)|\ge 3.$ Similarly $|D_y f(0,0)|\ge 4.$ Thus in $(1)$ we can solve for $g'(0)$ to see it is nonzero as desired.