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: A particular solution is $f(x)=x^2-1$. The general solution of the associated homogeneous problem
$$f\left({x\over3}\right)+f\left({2\over x}\right)=0$$
is $$f_{\rm hom}(x)=u\left(\log\bigl(\sqrt{3/2}\> x\bigr)\right)\qquad(x>0)\ ,$$
whereby $u$ is an arbitrary odd function. It follows that the general solution of the original problem is given by
$$f(x)=x^2-1+ u\left(\log\bigl(\sqrt{3/2}\> x\bigr)\right)\qquad(x>0)\ .$$
A: Suppose $f(x)$ can be given by the power series expansion $f(x) = a_{0} + a_{1} x + a_{2} x^{2} + \cdots$. Now, for the equation 
$$f\left(\frac{x}{3}\right) + f\left(\frac{2}{x}\right) = \frac{x^{2}}{9} - 2 + \frac{4}{x^{2}}$$
it is seen that:
\begin{align}
\frac{x^{2}}{9} + \frac{4}{x^{2}} - 2 &= 2 a_{0} + a_{1} \, \left(\frac{x}{3} + \frac{2}{x}\right) + a_{2} \left(\frac{x^{2}}{9} + \frac{4}{x^{2}}\right) + \cdots.
\end{align}
It is easy to determine that $a_{m+3} = 0$, for $m \geq 0$, $a_{1} = 0$, $a_{2} = 1$, and $a_{0} = -1$ which leads to the result
$$f(x) = x^{2} - 1.$$
A: 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
$$2f(\frac{\sqrt6}{3})=2(\frac{\sqrt6}{3})^2-2=2((\frac{\sqrt6}{3})^2+h)\Rightarrow h=-1$$
