You can't just lie in between a stranger and the first odd prime!

$f(x)+2f\left( \dfrac{2016}{x} \right) = 3x \quad \quad \small\text{1st Equation} \\ \text{substitute }x \rightarrow \dfrac{2016}{x} \\ f\left( \dfrac{2016}{x} \right) +2f(x)= 3\times \dfrac{2016}{x} \\ -2f\left( \dfrac{2016}{x} \right) -4f(x)= -6\times \dfrac{2016}{x} \quad \quad \small\text{2nd Equation} \\ \text{Add both the equations } \\ -3f(x)= 3x-6\times\dfrac{2016}{x} \\ f(x)=\dfrac{4032}{x}-x \\ \text{Now } f(144) = -116 \text{ and } f(72)=-16 \\ \text{Answer : } \dfrac{116}{16} = \boxed{7.25}$

\(\begin{array} {} x=72 & \implies f(72) + 2f(28) = 216 & ...(1) \\ x=28 & \implies f(28) + 2f(72) = 84 & ...(2) \\ x=144 & \implies f(144) + 2f(14) = 432 & ...(3) \\ x=14 & \implies f(14) + 2f(144) = 42 & ...(4) \end{array} \)

\(\begin{array} {} 2(2)-(1): & 3f(72) = -48 & \implies f(72) = - 16 \\ 2(4)-(3): & 3f(144) = -348 & \implies f(72) = - 116 \end{array} \)

$\implies \dfrac {f(144)}{f(72)} = \dfrac {-116}{-16} = \boxed{7.25}$

The title of this problem is hilarious. Solution: I just assumed the function took the form $f(x)=ax+\frac{b}{x}$ . If you substitute that into the functional equation, and equate the coefficients, you get a 2x2 linear system, which results in $f(x)=-x+\frac{4032}{x}$ .