Set $p:=f'(x)$ and
$$
\Phi(p)=\Phi_{a,b}:=\frac{ap}{\sqrt{1+ap^2}}+\frac{bp}{\sqrt{1+bp^2}}.
$$
The differential equation you wrote  can be rewritten as
$$
x=\Phi(p).
$$
If we could invert $\Phi$, then we could write
$$
f'(x)= p=\Phi^{-1}(x).
$$
For $a, b>0$ the function $\Phi$ seems to be increasing. The animation below depicts the graphs of $\Phi_{1,t}$ for $t\in [0,6]$

 [![enter image description here][1]][1]

It already shows that  the solution blows up in finite time. (Here I think of $x$ as time.)

The next animation depicts $\phi_{1,t}$  for $t=-1..0$ and you can see that the injectivity of $\Phi$ is lost  for some values of $t$.

[![enter image description here][2]][2]

**Remark.**  Here is an animation of the curve  described by Robert Bryant for $a=1$ and $b\in [-0.1,1]$, $t\in[-3,3]$

[![enter image description here][3]][3]


  [1]: https://i.sstatic.net/jpiUs.gif
  [2]: https://i.sstatic.net/4Msok.gif
  [3]: https://i.sstatic.net/egjYI.gif