let $f(\frac{x}{3})+f(\frac{2}{x})=\frac{4}{x^2}+\frac{x^2}{9}-2$ then find $f(x)$ let $f(\frac{x}{3})+f(\frac{2}{x})=\frac{4}{x^2}+\frac{x^2}{9}-2$
then find $f(x)$ 

My Try :
$$f(\frac{x}{3})+f(\frac{2}{x})=(\frac{2}{x})^2-1+(\frac{x}{3})^2-1$$
So we have :
$$f(x)=x^2-1$$
it is right ?Is there another answer?

A plausible $f(x)$ is in fact $f(x)=x^2-1$. A proof of that is as follows.
Suppose $f(x)=x^2+h$ For $x=\sqrt6$ one has $\dfrac 2x=\dfrac x3$ so
$$2f(\frac{\sqrt6}{3})=2(\frac{\sqrt6}{3})^2-2=2((\frac{\sqrt6}{3})^2+h)\Rightarrow h=-1$$
