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 $t\in [0,6]$

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 $t$.

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

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 $t\in [0,6]$

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 $t$.

Remark. Here is an animation of the curve described by Robert Bryant for $a=1$ and $b\in [-0.1,0.2]$, $t\in[-3,3]$

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 $t\in [0,6]$

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 $t$.

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 $t\in [0,6]$

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 $t$.

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