$f$ is continuous on $(0,+\infty)$ and
$$f^{\prime}(x)=\frac{(x^2)^{\prime}(1+x^2)-x^2(1+x^2)^{\prime}}{(1+x^2)^2}=\frac{2x+2x^3-2x^3}{(1+x^2)^2}=\frac{2x}{(1+x^2)^2}>0$$
so $f$ is increasing in $(0,\infty)$.
$$\lim_{x\to \infty}f(x)=\lim_{x\to \infty}\frac{x^2}{1+x^2}=\lim_{x\to \infty}\frac{1}{\frac{1}{x^2}+1}=1
$$ Therefore,
$$f((0,+\infty))=(f(0),\lim_{x\to \infty}f(x))=(0,1)$$
$f$ is thus surjective.
For the inverse: We know $f$ is injective in $(0,+\infty)$ and $f((0,+\infty))=(0,1)$.
$$f(x)=\frac{x^2}{1+x^2}\Leftrightarrow x^2f(x)+f(x)=x^2\Leftrightarrow x^2(1-f(x))=f(x)$$
Since $f(x)\neq 1$,
$$x^2=\frac{f(x)}{1-f(x)}$$
Since $x>0$,
$$x=\sqrt{\frac{f(x)}{1-f(x)}}$$
The inverse of $f$ is therefore,
$$f^{-1}(x)=\sqrt{\frac{x}{1-x}}$$