Consider a real function $f(r)$ with a real domain and ODE that has a form: $$ \Psi(f) \frac{\mathrm{d} f}{\mathrm{d} r}=\Phi(r,f) $$
where $\Psi$ and $\Phi$ are just some real functions that represent the left and right-hand sides. $\Psi$ can be e.g. $1-f$. If $f_0$ is such that $\Psi(f_0)=0$, e.g. $f_0$ = 1, I want to explore the ODE about the point $f_0=1$ and so I set $f = 1 + y$ and $r = \tilde{r} + x$ where $y$ is small compared to $1$ and $x$ is small compared to $\tilde{r}$ (I want to consider only linearly small terms . $\tilde{r}$ is such that $\Phi(\tilde{r},f(\tilde{r}))=0$ and so $\tilde{r}(f)$.
Dealing with the functions $\Psi$ and $\Phi$ is just Taylor expansion. However, I am a bit unsure how to understand the way to deal with point $\tilde{r}$ about which to expand the $r$. Specially, the derivative, on one hand:
$$ \frac{\mathrm{d} f}{\mathrm{d} r} = \frac{\mathrm{d} y}{\mathrm{d} r} = \frac{\mathrm{d} y}{\mathrm{d} x}\frac{\mathrm{d} x}{\mathrm{d} r} = \frac{\mathrm{d} y}{\mathrm{d} x}\left(1-\frac{\mathrm{d} \tilde{r}}{\mathrm{d} r}\right) = \frac{\mathrm{d} y}{\mathrm{d} x}\left(1-\frac{\mathrm{d} \tilde{r}}{\mathrm{d} f}\frac{\mathrm{d} f}{\mathrm{d} r}\right) $$ Which leads to: $$ \frac{\mathrm{d} f}{\mathrm{d} r} = \frac{\frac{\mathrm{d} y}{\mathrm{d} x}}{\frac{1+\mathrm{d} \tilde{r}}{\mathrm{d} f}\frac{\mathrm{d} y}{\mathrm{d} x}} $$ and plugging back in the ODE: $$ \Psi(1+y)\frac{\mathrm{d} y}{\mathrm{d} x}=\Phi(\tilde{r}+x,1+y)\left(1+\frac{\mathrm{d} \tilde{r}}{\mathrm{d} f}\frac{\mathrm{d} y}{\mathrm{d} x}\right) $$
On the other hand, since we are expanding about $f_0=1$ which happens for such $r=\tilde{r}$ when $\Phi(\tilde{r},f(\tilde{r})) = 0$ (which gives $\tilde{r}(f)$), and specifically $\Phi(\tilde{r}_0,1) = 0$, I am tempted to make the expansion of $r = \tilde{r}_0 + x$ where $\tilde{r}_0 = \tilde{r}(1)$ , that is it is a fixed point where $f(\tilde{r}_0)\equiv1$ and the derivative is immediately just $df/dr = dy/dx$.
I somehow feel that this second way is not correct as I am enforcing "something" but I cannot wrap my head around, why the second way is not correct, if $f=1$ when $r = \tilde{r}_0$, why do I have to still consider expanding $r$ about a variable point $r = \tilde{r} + x$?
I apologise, if the question is not clear but I did my best to explain the confusion.