Let's assume an ODE of the type $$y'=g(x)h(y(x)),$$ where $g:I\to\mathbb{R}$, $h:J\to\mathbb{R}$ are continuous functions and ${\xi\choose\eta}\in I\times J$.
In the case of $h(\eta)=0$ the solution is clear. So let's assume $h(\eta)\neq 0$. Then, we know that by solving the equation:
$$ \int_{\eta}^y\frac{1}{h(s)}ds=\int_{\xi}^xg(t)dt $$ with respect to $y$ delivers a solution of the ODE.
In our lecture the professor has proven this statement by constructing another function, namely
$$H(u):=\int_{\eta}^u\frac{1}{h(s)}ds$$ and then showing its invertibility. After some additional steps he finally comes to the conclusion that $$y:=H^{-1}\left(\int_{\xi}^xg(t)dt\right)$$ is a (unique) solution of the ODE.
In the literature this seems a standard way of proving the theorem. However, I was wondering why we cannot simply say that in the case of $h(\eta)\neq 0$ it directly follows from the implicit function theorem?
EDIT:
Some more details regarding the implicit function theorem:
Let be $y(\xi)=\eta$ and $F:\mathbb{R}^2\to\mathbb{R}$, where $$ F(x,y)=\int_{\eta}^y\frac{1}{h(s)}ds-\int_{\xi}^xg(t)dt. $$ Now we look at $F(x,y)\overset{!}{=}0$ and see that $F(\xi,y(\xi))=0$. All assumptions of the implicit function theorem are satisfied (I have skipped the details), so we know there exists a function $f$ such that $F(x,f(x))=0$. Further, we know how the derivative of $f$ looks like, namely $f'(x)= (-1) \left(\left(\int_{\eta}^{f(x)}\frac{1}{h(s)}ds\right)'\right)^{-1}\left(-\int_{\xi}^xg(t)dt\right)'=f(x) h(f(x)).$
Hence, the implicit function $f$ is a solution of the ODE. (I know this is a bit sloppy but I hope it explains the main idea)