Suppose I have a function $f: \mathbb{R^2} \rightarrow \mathbb{R}$, let's call it $f(x,y)$. Suppose I have a zero for the function, $(0,0)$ for example, at which one of the partial derivatives (which are continuous btw), vanish, say $f\_x$, but t'other, $f\_y$, doesn't. 

As I recall, the implicit function theorem (IFT) states I can get a graph $y=h(x)$ say, for the level set at height zero such that $f(x,h(x))=0$ in some open neighbourhood $x \in (-\epsilon,\epsilon)$. What I am wondering is if, with a few parsimonious other details on $h(x)$, can I get sufficient conditions so that over a possibly small interval, $h^{-1}$ exists, so that in fact $f(h^{-1}(y),y)=0$ over some restricted interval for $y$, even if $f\_x(0,0)=0$ which rules out applying the IFT in this direction. My naive approach is try and establish monotonicity of $h$ on some suitable interval...

Now the IFT does say that $h(x)$ is differentiable (therefore continuous) on $(-\epsilon,\epsilon)$ with $h'(x)=-f\_x(x,h(x))/f\_y(x,h(x))$.

I think that if I further add the conditions that $h(x)$ is **continuously** differentiable and not constant, then this might be enough, since geometric intuition (health warning!) leads me to believe that this means that there is some subinterval $I \subset (-\epsilon,\epsilon)$ on which $h(x)$ is monotone. In this case I think that this means there must be a possibly small interval $[0,\delta)$, (the left bound suggested by the fact that $h(0)=0$) on which $h$ is monotone. The questions therefore is, is this rigmarole at all likely to succeed or doomed to fail? 

Adding the proviso that $h$ is continuously differentiable perhaps precludes the 'everywhere differentiable but nowhere monotone' counter examples discussed in [this question on MO](https://mathoverflow.net/questions/167323/everywhere-differentiable-function-that-is-nowhere-monotonic). I do not have good enough analysis references immediately to hand on this, if anyone can recommend something I'd be grateful. 

I'm obviously trying to get as far as I can get to say something about the level set in the $y$ direction without invoking too many specifics about the function $f(x,y)$ itself. So in short, my $h(x)$ has a zero at $x=0$, it's continuously differentiable and not constant, surely that must mean I can find a $\delta$ such that it's monotone on $[0,\delta)$, which means I can say something about the level set in the $y$-direction of $f(x,y)$ even if the full IFT cannot nail it. Effectively I'd like a one-sided IFT in one direction given nice enough $h(x)$. My analysis is very rusty, the above approach is ungainly and probably there are much better ways to explore the level sets if a partial derivative vanishes than trying to establish piecewise monotonicity using continuous differentiability of $h(x)$ (the detail I am allowing), but would be grateful for any pointers, thanks!