It is given ,
$\large f(x)$ is strictly decreasing function so ,
$\large f'(x) <0$ for any x in $\large \left ( 0,\infty \right )$.
$\large h(x)=\frac{f(\frac{x}{x+1})}{1+f(\frac{x}{x+1})}$
$\large h(x)=1-\frac{1}{1+f(\frac{x}{x+1})}$
$\large h'(x)=-\frac{(-1)f'(\frac{x}{x+1})\frac{\mathrm{d} }{\mathrm{d} x}(\frac{x}{x+1})}{(1+f(\frac{x}{x+1}))^{2}}$
$\large h'(x)=\frac{f'(\frac{x}{x+1})\frac{1}{(1+x)^{2}}}{(1+f(\frac{x}{x+1}))^{2}}$
now here , $\large (1+f(\frac{x}{x+1}))^{2}>0$,$\large (\frac{1}{(x+1)^{2}})>0$
so now, the sign of $\large h'(x)$ is only depends on $\large f'(\frac{x}{x+1})$.
as it is given initially $\large f(x)$ is strictly decreasing function so $\large f'(x)$ is always less than zero .
so ,$\large f'(\frac{x}{x+1})<0$ .
so , $\large h'(x)<0$ for any x in $\large \left ( 0,\infty \right )$
so ,$\large h(x)$ is strictly decreasing .
so correct option is (A).