We consider the following O.D.E \begin{align}(a)\qquad\qquad\begin{cases}x'(t)=\dfrac{\rho^2(x(t))}{1+\rho^2(x(t))} & t\geq 0,\\x(0)=x_0\in \Bbb{R}&\end{cases}\end{align} where \begin{align}\rho:\Bbb{R}\to \Bbb{R}\end{align} is a $C^1$ function. I want to prove that $(a)$ has a unique solution defined on $\Bbb{R}^+?$
CURRENT PROOF
Since $\rho:\Bbb{R}\to \Bbb{R}$ is $C^1$ function, $\rho^2$ is also a $C^1$ function. Also, \begin{align}\dfrac{\rho^2}{1+\rho^2}\end{align} is a $C^1$ function. \begin{align}\rho^2(x(t))\leq 1+\rho^2(x(t)),\;\;\forall\,t \geq 0,\;x\in \Bbb{R}\end{align} \begin{align}f(x(t))=\dfrac{\rho^2(x(t))}{1+\rho^2(x(t))} \leq 1,\;\;\forall\,t \geq 0,\;x\in \Bbb{R}\end{align}
This implies that $f$ is affine. So, $(a)$ has a unique maximal solution on $[0,t_{\max})$ and $t_{\max}=+\infty$. Therefore, $(a)$ has a unique solution defined on $\Bbb{R}^+?$
Kindly confirm if this proof is fine or not. If no, alternative proofs will be accepted. Thanks!