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 $f$:

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