A logarithm-like functional equation

Suppose we are given that a monotonically decreasing smooth function $f$ on $(0,\infty)$ obeys the functional equation $f(x) = -f(\frac{1}{x})$, and satisfies $f(\frac{1}{3}) = \frac{1}{2}$ and $f(\frac{1}{2}) = \frac{1}{3}$. Furthermore, $\lim\limits_{x\rightarrow0} f(x) = 1$. Is there a way to infer information about the function from these data alone, or even classify all functions satisfying them? I see that a function proportional to $\log x$ satisfies the functional equation, but cannot satisfy the special values.

I now found a function satisfying these data: $f(x) = \dfrac{1-x}{1+x}$.

The following Mobius function has the desired properties you want:

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

$$f(0)=1,f(x)=-f \left(\frac{1}{x}\right),f \left(\frac{1}{3} \right)=\frac{1}{2},f \left(\frac{1}{2} \right)=\frac{1}{3}$$

EDIT:

I did not pay attention to the modification of the original question that added the right answer while I was typing my answer to the original question above.

Here I post a general method to deal with such problem by providing a new function $g(x)$ which satisfies the desired the requirements.

Define $$g(x):=\frac{a(x-x^{-1})+b(x-x^{-1})^3+c(x-x^{-1})^5}{u x^5+v x^{-5}}$$

Then

$$g(x)+g(x^{-1})=0 \text{ requires that } u=v$$.

$$\lim_{x\to 0}g(x)=-\frac{c}{v}=1 \text{ requires that } v=-c$$

$$\lim_{x\to \infty}g(x)=\frac{c}{v}=-1 \text{ requires that } v=-c$$

We can then solve $$g(1/3)=\frac{1}{2}\text { and } g(1/2)=\frac{1}{3}$$ for $a$ and $b$ and obtain:

$$a=\frac{2683}{504}c \text{ and } b=-\frac{61}{42}c \text{ }(c\not = 0)$$