A simple geometric problem, solving $f'(x)=\frac{f(x)}{\sqrt{r(x)^2-f(x)^2}}$, given $r(x)$.

Here we assume in general that the plane curve is $C\in\textbf{C}^{(1)}[a,b]$ and of the form $$ C:\overline{x}(s)=\left\{A(s),B(s)\right\},\tag 1 $$ where $s$ is its canonical parameter i.e. $$ (A'(s))^2+(B'(s))^2=1.\tag 2 $$ If $P_0\in C$ (a point of the curve) and $t(P_0)$ is the tangent line in $P_0$ and $O_1=(x_1,0)$ is the intersection of $t(P_0)$ with the $x-$axis, then $r(s_0)=|P_0O_1|$. Hence $$ t(P_0):y-B(s_0)=\frac{B'(s_0)}{A'(s_0)}(x-A(s_0)) $$ and $O_1:y_1=0$. Hence
$$ -B(s_0)=\frac{B'(s_0)}{A'(s_0)}(x_1-A(s_0))\Leftrightarrow x_1-x_0=-B(s_0)\frac{A'(s_0)}{B'(s_0)}. $$ $$ r(s_0)=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}=\sqrt{B(s_0)^2\frac{(A'(s_0))^2}{(B'(s_0))^2}+(B(s_0)^2)}\Leftrightarrow $$ $$ r(s_0)=\frac{|B(s_0)|}{|B'(s_0)|}\sqrt{(A'(s_0))^2+(B'(s_0))^2}=\left|\frac{B(s_0)}{B'(s_0)}\right|. $$ Hence when $s_0$ varies, we get $$ \frac{B'(s)}{B(s)}=\frac{\pm 1}{r(s)}\Rightarrow B(s)=\exp\left(\pm\int\frac{ds}{r(s)}\right). $$ Also form (2) we get $$ A(s)=\pm\int\sqrt{1-(B'(s))^2}ds=\pm\int\sqrt{1-\frac{B(s)^2}{r(s)^2}}ds= $$ $$ =\pm\int\sqrt{1-\frac{\exp\left(\pm2\int r^{-1}(s)ds\right)}{r(s)^2}}ds $$ Hence for a given $r(s)$, the curve can be parametrized as $$ \overline{x}(s)=\left\{\pm\int\sqrt{1-\frac{1}{r(s)^2}\exp\left(\pm2\int\frac{ds}{r(s)}\right)}ds,\exp\left(\pm\int\frac{ds}{r(s)}\right)\right\},\tag 3 $$ for $s\in[a,b]$ and $s$ being its canonical parameter.

Solve the following ODE for $f(x)$ knowing $r(x)$: $$f'(x)=\frac{f(x)}{\sqrt{r(x)^2-f(x)^2}}$$

I tried to simplify this equation by different substitutions.

1) Multiplying both its sides by $f(x)$ and putting $g(x)=f(x)^2$, we obtain $$g’(x)=\frac{2g(x)}{\sqrt{r(x)^2-g(x)}}.$$

2) Putting $h(x)= r(x)^2-g(x)$, we obtain

$$2r(x)r’(x)-h’(x)=\frac{2 r(x)^2-2h(x)}{\sqrt{h(x)}},$$ $$h’(x)=2\left({\sqrt{h(x)}}-\frac{r(x)^2}{\sqrt{h(x)}}+r(x)r’(x)\right).$$

3) Putting $v(x)=\sqrt{r(x)^2-f(x)^2}$, we obtain

$$v’(x)=\frac{r(x)r’(x)-f(x)f’(x)}{v(x)}= \frac{r(x)r’(x)-\tfrac{f(x)^2}{v(x)}}{v(x)}=$$ $$\frac{r(x)r’(x)-\tfrac{r(x)^2-v(x)^2}{v(x)}}{v(x)}= 1+\frac{r(x)r’(x)}{v(x)}-\frac{r(x)^2}{v(x)^2}.$$

4) Putting $u(x)=\frac 1{v(x)}$, we obtain

$$-\frac{u’(x)}{u(x)^2}=1+r(x)r’(x)u(x)-r(x)^2u(x)^2,$$

$$u’(x)= r(x)^2u(x)^4-u(x)^2-r(x)r’(x)u(x)^3.$$

We can try to solve the last two equations in terms of power series for $r$, $u$, and $v$.

Just another idea. Have you tried to look for parametrized solutions? That is, starting with $$r(x)^2=f(x)^2\left(1+\frac{1}{f'(x)^2}\right)$$ and instead of looking for solutions $(x,f(x))$ you reparametrize $x(t)=h(t)$ for some function $h$. In particular, for example, assuming $f(x)$ is invertible on some domain and choosing $h(t)=f^{-1}(t)$, you then obtain the ode for $x(t)$, so $$r(x)^2 = t^2 \left(1+x'(t)^2\right) \, .$$

The solution would then be given by $(x(t),t)$ for some range of the parameter $t$. Unfortunately, except for the simple case $r(x)={\rm const}$, this still does not allow separation of variables, but maybe you can work from there with a different ansatz for $h(t)$.

This is just offering a different approach, which produce a different ode.
Let's put the problem in the Tractrix perspective.

Consider the two reference systems, one fixed and one attached the "tractor" at $\xi$ and which is moving at constant speed $v$ along the $x$ axis.

Assume that we know $r$ as function of $\xi$.

The speed of $P$ in the moving frame has the polar components $r', \, r \alpha '$, and in the fixed frame $v$ is added to those.
The resultant speed normal to the curve and thus to $r$ shall be null, so $$ \left\{ \matrix{ r\alpha ' = v\cos \left( {\alpha - \pi /2} \right) = v\sin \alpha \hfill \cr r = r(\xi ) = r(vt) \hfill \cr} \right. \tag {1}$$ we can rewrite the first expression separating the variables $$ {{d\alpha } \over {\sin \alpha }} = {v \over {r(vt)}}dt $$ which means $$ {{d\alpha } \over {\sin \alpha }} = {1 \over 2}d\ln \left( {{{1 - \cos \alpha } \over {1 + \cos \alpha }}} \right) = d\ln \left( {\tan \left( {\alpha /2} \right)} \right) = {v \over {r(vt)}}dt = {{d\xi } \over {r(\xi )}} $$

So under the assumption of $r=r( \xi )$, we arrive at a closed parametric equation for the curve as seen from the moving frame $$ \left\{ \matrix{ d\ln \left( {\tan \left( {\alpha /2} \right)} \right) = {{d\xi } \over {r(\xi )}} \hfill \cr r = r(\xi ) \hfill \cr} \right. \tag {2}$$ (of course provided that the integral of $ {{d\xi } \over {r(\xi )}}$ be known in closed form).