Solve $f(x)+f(1+\frac{1}{x})=1+x$ The title means

Does there exists any function $f:\Bbb R\to\Bbb R$ s.t. it satisfies the following equation?
$$f(x) + f(1+\frac1x)=1+x\quad\text{for all } x\not=0.$$

This question is made by my friend. He does not restrict the function with any further condition, though adding some condition may be necessary.
I found some similar questions, e.g. $f(x)+f(1-\frac1x)=1+x$. For this example, I could replace $x$ by $1-\frac1x$ and $\frac1{1-x}$, then just solving the system of linear equations. Unfortunately, these operations can't be used for present question.
 A: Preliminaries
Consider the two-ended sequence of extended reals:
$$\{\cdots,-\frac23,-\frac12,\overbrace{-1}^{a_{-1}},\overbrace{0}^{a_0},\overbrace{\infty}^{a_1},1,2,\frac32,\frac53,\ldots\}$$
where $a_{n+1}=1+\frac{1}{a_n}$. Note that the function $g(x)=1+\frac1x$ is a bijection from $(a_n,a_{n+2})$ to $(a_{n+1},a_{n+3})$, with inverse $g^{-1}(x)=\frac{1}{x-1}$. We have to abuse interval notation for this and rearrange the order of the interval endpoints as needed. For example $$(a_2,a_4)\mapsto(a_5,a_3)$$
$$(1,3/2)\mapsto(5/3,2)$$
Also, the symbol $\infty$ may mean $\infty$ or it may mean $-\infty$. We have
$$(-\infty,-1)\mapsto(0,1)\mapsto(2,\infty)$$
And overall:
$$\cdots\mapsto(-1,-2/3)\mapsto(-1/2,0)\mapsto(-\infty,-1)\mapsto(0,1)\mapsto(2,\infty)\mapsto(1,3/2)\mapsto(5/3,2)\mapsto\cdots$$
Again, these are bijections.
$f$ is Arbitrary on $[0,1]$ and $f(x)=\frac{1+x}2$ if $x=\phi, -1/\phi$
Define $f$ to be any function on $(0,1)$, and then the relation forces the value of $f$ on all of the rest of these intervals according to:
$$f(g(x))=1+x-f(x)$$
$$f(g^{-1}(x))=1+g^{-1}(x)-f(x)$$
So the value of $f$ on $(0,1)$ determines the value of $f$ on $$\cdots\cup(-1,-2/3)\cup(-1/2,0)\cup(-\infty,-1)\cup(0,1)\cup(2,\infty)\cup(1,3/2)\cup(5/3,2)\cup\cdots$$
which is more nicely written as
$$\overbrace{(-\infty,-1)\cup\cdots\cup(-1,-2/3)\cup\cdots}^{\text{right endpoints converging to $(1-\sqrt{5})/2$}}\cup\cdots(-1/2,0)\cup(0,1)\cup(1,3/2)\cup\overbrace{\cdots\cup(5/3,2)\cup(2,\infty)}^{\text{left endpoints converging to $(1+\sqrt{5})/2$}}$$
It remains to explore the possibilities for $f$ at the endpoints of these intervals and their two limit points. Define a value for $f(a_2)=f(1)$, and that implies a value for $f(a_3)=f(2)$, $f(a_4)=f(3/2)$ and so on. The recurrence relation $x\mapsto g(x)$ implies the limit of these values of $x$ will be $\phi=\frac{1+\sqrt{5}}{2}$. Similarly, define a value for $f(a_0)=f(0)$, and that implies a value for $f(a_{-1})=f(-1)$, $f(a_{-2})=f(-1/2)$ and so on. The recurrence relation $x\mapsto g^{-1}(x)$ implies the limit of these values of $x$ will be $-1/\phi$.
Substituting $\phi$ or $-1/\phi$ for $x$ in the functional equation, we find that $f(\phi)=\frac{1+\phi}2$ and $f(-1/\phi)=\frac{1-1/\phi}2$.
Note that $g(0)$ is undefined ($\infty$), as is $g^{-1}(1)$. So the bridge from $0$ to $1$ is broken. For example a value for $f(0)$ would not imply a value for $f(1)$. However if $f$ is extended to $\mathbb{R}\cup\{\infty\}$, then indeed a value chosen for $f(0)$ would imply a value for $f(1)$.
Note however that continuity over $\mathbb{R}$ is not possible. A (finite) value for $f$ at $0$ implies a finite limit at $\infty$, which in turn implies an infinite (non)value for $f$ at $1$.
Here is a picture of one such function seeded by $2x-\frac1x$ on $(0,1)$. This happens to make the intervals continuously meet up for the most part. Only six intervals are shown.

