$\text{ Let }f\text{ be a function satisfying }f\left(xy\right)=\frac{f\left(x\right)}{y}\text{ for all positive real numbers }x\text{ and }y\text{ . If }$$f\left(30\right)=20\text{ , then the value of }f\left(40\right)\text{ is }$

An equation of this kind is called a functional equation, and can often be solved by choosing particular values for the variables. 

In this case, by choosing we see that 

$\begin{array}{l}f\left(y\right)=\frac{f\left(1\right)}{y}\text{ for all }y.\text{ Put }y=30;f\left(1\right)=30.f\left(30\right)=30.20=600.\\ \text{ Now }f\left(40\right)=\frac{f\left(1\right)}{40}=\frac{600}{40}=15\end{array}$