Functional Equation with Inverse

How do I solve the following functional equation:

$$f(x)+12f^{-1}(x)=\frac{1}{x}f(x)$$

I've been doing a lot of functional equations, but I haven't done one yet that has the function and its inverse together. All I've done so far is figure out that $f(x)$ has a fixed point at $x=\frac{1}{13}$ and that $f(0)$ starts a cycle of orbit $2$.

Thanks! All help is appreciated!

Answer

We have

$$y=f(x)$$

and:

$$x=f^{-1}(y)=g(y)$$

Thus

$$(\frac{1}{x}-1)f(x)=12f^{-1}(x)\\ x=f(f^{-1}(x))=f(\frac{1}{12}(\frac{1}{x}-1)f(x))\\ $$

Thus, we can formulate a recursive numeric search on $f$:

$$min_{df}\left|x-f\left(\frac{1}{12}(\frac{1}{x}-1)f(x)+df(x)\right)+df\left(\frac{1}{12}(\frac{1}{x}-1)f(x)\right)\right|\\ $$