Let  f be a real valued function such that $f\left(x\right)+2f\left(\frac{2002}{x}\right)=3x\forall x>0$ then the value of  $\frac{f\left(2\right)}{2000}$ is

Substitute $x=2$ in the given equation $f\left(2\right)+f\left(1001\right)=3\left(2\right)=6$

Now substitute $x=1001$

Hence, $f\left(1001\right)+2f\left(2\right)=3003$
Solving the above two equations we get $f\left(2\right)=2000$

Therefore, $\frac{f\left(2\right)}{2000}=1$