Local implicit function theorem and Global implicit function theorem

Question

Suppose that $f=f(x,y)$ is $C^1$ in $x$ and $y$, where $x,y\in\mathbb{R}$ and $f(x_0,y_0)=0$. Moreover, $\frac{\partial f(x,y)}{\partial y}>0$ for all $x,y\in\mathbb{R}$. Could we conclude that $y$ can be solved implicitly from $f(x,y)=0$ in terms of $x$ as $y=y(x)$ for all $x\in\mathbb{R}$? If the answer is no and an additional assumption is imposed, i.e. we can obtain $y'(x)>0$ for all $x\in\mathbb{R}$ by implicitly differentiating $f(x,y(x))=0$ with respect to $x$, then these assumptions are enough to get the same conclusion?

Answer 1

I'm not sure what you want to know beyond what is shown in this question and it's answer, but that question shows already where you have to expect the result to break down. As an example, consider

$$F(x,y) = - e^{-y} +x$$ Clearly, $\frac{\partial F}{\partial y} = e^{-y}>0$ everywhere, but it is easy to see that $F$ is strictly negative for $x\le0$ on the one hand side but admits a (obviously unique) solution of $F(x,y) = 0$ for each $x>0$. This shows that if the bounds on the derivates no longer hold, the solution may go off to infinity (Which is exactly what happens in this example).