Here,  $f:\left(0, \infty \right)\to \left(0, \infty \right),$ i.e. domain is $(0, \infty )$ and codomain is $\left(0, \infty \right).$

For one-one, $f\left(x\right)=\frac{x}{1+x}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{{\left(1+x\right)}^2}>0, \forall x\in \left(0, \infty \right)$

$\therefore f\left(x\right)$ is increasing in its domain. Thus, $f\left(x\right)$ is one-one in its domain.

For onto (we find range)

$f\left(x\right)=\frac{x}{1+x} \Rightarrow y=\frac{x}{1+x} \Rightarrow y+yx=x$

$\Rightarrow x=\frac{y}{1-y} \Rightarrow \frac{y}{1-y}\ge 0$ as $x\ge 0$ $\therefore 0\le y<1$

i.e., Range$\ne$Codomain

$\therefore f\left(x\right)$ is one-one but not onto.