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

Solution 1:

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

Solution 2:

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 $$f(x) = x^{2} - 1.$$