Find a non constant function that is a quotient of two polynomial, for which: $f\left(x+\frac{1}{f(x)}\right)=f(x)f(-x)$ Before I post the problem, I want to ask what is wrong with the exactly same problem I posted three days ago, 'cause nobody seemed willing to answer it.
The non constant function must satisfy the equation:
$$f\left(x+\frac{1}{f(x)}\right)=f(x)f(-x).$$
The function could also be written as: $f[x(x+\frac{x}{xf(x)})]$
 A: Edited. My conjecture is as follows:

Conjecture. The only non-constant solutions are of the form $ f(x) = 1/(x^{2} - x + c) $ for some $c \in \Bbb{C}$.

By plugging $g(x) = 1/f(x)$ instead, the functional equation reduces to
$$ g(x+g(x)) = g(x)g(-x). \tag{1} $$
Step 1. Since $\text{(1)}$ is true for any real $x$, by the Identity Theorem this relation continues to hold on the Riemann sphere $\hat{\Bbb{C}} = \Bbb{C} \cup \{\infty\}$, where $g$ is holomorphic as mapping $\hat{\Bbb{C}} \to \hat{\Bbb{C}}$. What we want to prove first is as follows:

Claim. $g(\infty) = \infty$.

Assume that $g(\infty) \neq \infty$. Since $g$ is non-constant, the Liouville's Theorem tells us that $g$ must have a pole at some $x_{0} \in \Bbb{C}$. Then $g(x_{0}) = \infty$ shows that
$$ g(-x_{0})= \frac{g(x_{0}+g(x_{0}))}{g(x_{0})} = \frac{g(\infty)}{\infty} = 0 $$
and hence $g(x) = (x+x_{0})^{m}h(x)$ for some $m \geq 1$ and rational $h$ with $h(-x_{0}) \neq 0$. Thus in view of the relation $\text{(1)}$, we have
\begin{align*}
g(-x)
 = \frac{g(x+g(x))}{g(x)}
&= \frac{(x+x_{0}+(x+x_{0})^{m}h(x))^{m}h(x+g(x))}{(x+x_{0})^{m}h(x)} \\
&= \frac{(1+(x+x_{0})^{m-1}h(x))^{m}h(x+g(x))}{h(x)}.
\end{align*}
Taking $x \to -x_{0}$ we obtain $g(x_{0}) \neq \infty$, which contradicts our assumption. Therefore the claim follows.
Step 2. Now we have two cases:


*

*$x + g(x) \to \infty$ as $x \to \infty$.

*$x + g(x) \to \alpha$ for some $\alpha \in \Bbb{C}$ as $x \to \infty$. In other words, $g(x) = -x + \alpha + h(x)$ for some $h(x)$ vanishing at $x = \infty$.


We consider the case 1. Our claim is the the conjecture is true in this case.

Claim. $g(x)$ is of the form $x^{2} - x + c$ in the case 1.

By the Step 1, $g(x) \sim A x^{d}$ as $x \to \infty$ for some $d \geq 1$ and $A \neq 0$. Then the assertion of the case 1 says that $x + g(x) \sim A'x^{d}$ (where $A' = A$ if $d \geq 2$ and $A' = A+1 \neq 0$ if $d=1$). So we have
$$ AA'^{d}x^{d^{2}} \sim g(x+g(x)) = g(x)g(-x) \sim A^{2}(-1)^{d} x^{2d}. $$
Comparing both the exponent and the coefficient, we have $d = 2$ and $A = 1$. This allows us to expand $g$ as the Laurent series at $x = \infty$ by
$$ g(x) = x^{2} + bx + c + \sum_{n=1}^{\infty} \frac{a_{n}}{x^{n}} \quad \text{as } x \to \infty. $$
Plugging this to $\text{(1)}$ and comparing both sides up to degree 3, we have
$$ 0 = g(x+g(x)) - g(x)g(-x) = 2(b+1)x^{3} + \mathcal{O}(x^{2}) \quad \text{as } x \to \infty $$
and hence $b = -1$. So it remains to show that $a_{n} = 0$ for all $n \geq 1$. To this end, assume otherwise and let $m$ be the smallest positive integer for which $a_{m} \neq 0$. Then we can write
$$ g(x) = x^{2} - x + c + \frac{h(x)}{x^{m}}, \quad \text{where} \quad h(x) = \sum_{n=0}^{\infty} \frac{a_{n+m}}{x^{n}}. $$
Then $h(x)$ is a rational function with $h(\infty) = a_{m}$ and plugging this to $\text{(1)}$,
\begin{align*}
0
&= x^{2n}\{ g(x+g(x)) - g(x)g(-x) \} \\
&= (x^{2} - x + c)\{ h(x) - (-1)^{m} h(-x) \} - h(x) + \mathcal{O}(x^{-1}).
\end{align*}
Expanding the last expression and comparing the coefficients shows that in any cases we must have $a_{m} = 0$, a contradiction! Therefore $a_{n} = 0$ for any $n$ and the conjecture is true in the case 1. ////
Unfortunately, I have no idea how to deal with the exceptional case 2.
