Let $f: \mathbb{R} \to \mathbb{R}$ and consider the problem of solving $f(x) = y$.
The inverse function theorem says $$ \mathrm{d} x= \frac{\mathrm{d} y}{f'(x)} $$
We could turn this ODE into a finite difference scheme: $$x_{n+1} - x_{n} = \frac{\Delta y}{f'(x_n)}$$
where $\operatorname{sgn}(\Delta y)$ = $\operatorname{sgn}(y - y_n)$ in order to guide $y_n$ towards $y$.
If the stepsize is $|\Delta y| = 1$, we get Newton's method: $$ x_{n+1} - x_n = \frac{y - f(x_n)}{f'(x_n)} $$
Is there more to this correspondence or is the above the extent of it?
First, what does $dx=dy/f'(x)$ really mean? If you are treating differentials as finite differences, then this equation is really an approximation and rearranging it gives $dy/dx=f'(x)$. But this just says the derivative is (approximately) the slope of the secant line...which isn't saying much (and is not the inverse function theorem).
This is the problem with using differential notation...it allows for such heuristic hand-waving.
Since you are really starting with a discrete approximation for the derivative, your heuristic manipulations wind up at Newton's method, since Newton's method is essentially based on approximating a secant line (one that intersects the actual root of a function) with a tangent line.
In particular, given constant $y$, define $g(x):=y-f(x).$ We wish to find a root of $g$. Pick some $x_n$ close enough to the root. We hope $x_{n+1}$ will get us closer to the root. To do so, we choose $x_{n+1}$ as the root of the tangent line to $g(x)$ at $x=x_n$:
$$g'(x_n)= \frac{g(x_n)-0}{x_n-x_{n+1}}\implies x_{n+1}-x_n=-\frac{g(x_n)}{g'(x_n)}=\frac{y-f(x_n)}{f'(x_n)},$$
which is the Newton's method algorithm you obtained.