Let $f: U \rightarrow \mathbb{R}$ be continuous on an open set $U \subseteq \mathbb{R}^2$, differentiable at a point $(a,b) \in U$, where $\partial_{y} f(a,b) > 0$ and $f(a,b) = c$. Prove that there exists an interval $I$ and a function $\xi: I \rightarrow J$ such that $\xi(a) = b$ and, for every $x \in I$, $f(x, \xi(x)) = c$. Show also that $\xi$ is differentiable at $a \in I$.
I can't solve the last part of the exercise above. The existence of $\xi$ follows a similar idea to the proof of the Implicit Function Theorem. However, differentiability of $\xi$ seems to be a bit tricky. Here's what I tried.
Sketch: It's plausible that the derivative of $\xi$ at $a \in I$ is $-\partial_x f(a,b)/\partial_y f(a,b)$, analogous to the IFT, so we will try to bound the following expression: $$\left|\frac{\xi(a+h) - \xi(a)}{h} - \left(-\frac{\partial_x f(a,b)}{\partial_y f(a,b)}\right)\right| = \left|\frac{\partial_y f(a,b) (\xi(a+h) - \xi(a)) + \partial_x f(a,b) h}{\partial_y f(a,b) h}\right|.$$
We want to rewrite some of those terms above. Define $k(h) := \xi(a+h) - \xi(a)$ (I'll just use $k$ for convenience). Then, given that $f$ is differentiable, there exists an error function, $$r(h,k) = f(a+h, b+k) - f(a,b) - \partial_x f(a,b) h - \partial_y f(a,b) k,$$ such that $r(h,k)/|(h,k)| \rightarrow 0$ as $|(h,k)| \rightarrow 0$. Notice that $$b + k = \xi(a) + (\xi(a+h) - \xi(a)) = \xi(a+h).$$ Therefore, $f(a+h, b+k) = f(a+h, \xi(a+h)) = c$ and $f(a+h, b+k) - f(a, b) = 0$. Now, our error function reduces to $$r(h,k) = - \partial_x f(a,b) h - \partial_y f(a,b) k,$$ which is exactly the numerator of the expression we had at the beginning. Substituting $r(h,k)$ there, we have $$\left|\frac{\xi(a+h) - \xi(a)}{h} - \left(-\frac{\partial_x f(a,b)}{\partial_y f(a,b)}\right)\right| = \left|\frac{1}{\partial_y f(a,b)}\frac{r(h,k)}{h}\right|.$$
Now, here's the issue: I'm not sure whether $r(h,k)/h$ goes to zero as $h$ goes to zero. I know $k$ goes to zero, since $\xi$ is continuous at $a \in I$. However, differentiability of $f$ only guarantees $r(h,k)/|(h,k)| \rightarrow 0$ as $|(h,k)| \rightarrow 0$. I tried rearranging the expression as
$$\frac{r(h,k)}{h}=\frac{r(h,k)}{|(h,k)|}\frac{|(h,k)|}{h},$$ but I'm not even sure whether $|(h,k)|/h$ is bounded or not. Does anybody have any idea how to proceed?