<s>[INST] 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?
 [/INST] 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
$$f(x)=x^2-1+ u\left(\log\bigl(\sqrt{3/2}\> x\bigr)\right)\qquad(x>0)\ .$$
</s>