Implicit function, not obvious version I'm not sure if the title is meaningful.
Here is the problem:
Let $F: \mathbb{R}^2 \rightarrow \mathbb{R}$ be class $C^1$ and $F(0,0)=0$ and $\forall (x,y)\in \mathbb{R}^2: 0<|\frac{\partial F}{\partial x}(x,y)|<|\frac{\partial F}{\partial y}(x,y)|$
Prove that there exists $\varphi : \mathbb{R} \rightarrow \mathbb{R}$ class $C^1$
such that $\{ (x,y)\in \mathbb{R}^2 | F(x,y)=0\} = \{(x, \varphi(x) | x \in \mathbb{R})\}$
I have no idea how to approach this.
Could you help me?
Thank you!
 A: The implicit function theorem immediately implies that such a function $\varphi$ exists in a neighbourhood of the origin (is this clear to you?). 
The nonobvious parts are the claims that i) a global solution exists and ii) it is (globally) the only solution. 
i) something like this is usually done by showing that the interval on which you can solve is open and can always be extended to include it's boundary. The fact that it is open is, again, an immediate consequence of the implicit function theorem and the assumption that the derivative in $y$ direction does not vanish. So assume you have an interval $(-a, b), a, b >0$ and $\varphi$ is defined on $(-a, b)$ such that $F(x,\varphi(x)) = 0 $ along that interval.
The point to show i) is, that you can estimate the derivative of $\varphi$. This is true since
$$0 = \frac{d F}{d x} (x, \varphi(x)) =\left(\frac{\partial F}{\partial x}+ \frac{\partial F}{\partial y}\frac{d\varphi}{dx} \right)(x, \varphi(x))$$
and consequently (this is the key observation)
$$0 < |\frac{d\varphi}{dx} (x)| = \frac{|-\frac{\partial F}{\partial x}(x, \varphi(x))|}{\frac{\partial F}{\partial y}(x, \varphi(x))}<1$$
(because of the inequalities you were given). This implies that $\varphi$ can be extended in $C^1$ to the boundary of any finite interval (use the mean value theorem to show this), hence to $\mathbb{R}.$
As for ii), look at  $(x,\varphi(x)), x\in\mathbb{R}$. This is a $C^1$ embedded submanifold, and each point $(x,y)$ in the plane can be reached by drawing a line parallel to the $y$-axis from $(x,\varphi(x))$ to that point. Along such a line,   $F$ is strictly increasing if $y > \varphi(x)$ and decreasing if $y< \varphi(x)$ (again using the inequlites on the derivatives), hence it can be $0$ only in $(x, \varphi(x)).$ 
Edit (in response to a comment, show that $\varphi$ may be continuosly extended and a small correction): Any constant $C>0$ for which 
$$0 < |\frac{d\varphi}{dx} (x)| \le C $$
will do. The value $1$ in the above inequality results from the assumptions in the question. 
If $\varphi$ exists on $(a,b)$, say, one first has to show that it can be continuously extended up to the boundary, e.g. to, $b$. To see this consider any sequence $x_k \rightarrow b$. Then $|\varphi(x_k)-\varphi(x_l)| = |d\varphi(\xi)||x_k-x_l| $ by the mean value theorem for some $\xi$, which depends on $x_k, x_l$. But if the derivative is bounded, then the rhs in this equality is bounded by $C|x_k - x_l|$ and since $(x_k)$ is Cauchy this implies that $\varphi(x_k)$ converges to a continuous extension $\varphi(b)$. Now note that in the 'key observation' you can omit the $|.|$ and have an equality with a continuous function on the rhs, i.e. 
$$\frac{d\varphi}{dx} (x) = \frac{-\frac{\partial F}{\partial x}(x, \varphi(x))}{\frac{\partial F}{\partial y}(x, \varphi(x))}$$
Since the right hand side is continuos up to $b$ so is the left hand side.
