parmenides51 wrote: Find the function $\displaystyle{f : \mathbb{R}-\left\{ 0\,\right\} \rightarrow \mathbb{R} }$ such that $$f(x)+f(1-\frac{1}{x}) = \frac{1}{x}$$(-InnoXenT-) I suppose that missing domain of functional equation is $\forall x\in\mathbb R\setminus\{0\}$ If so : no such function (just set $x=1$ and second summand of LHS is undefined)

I just added domain for the equation, $x \in \mathbb{R}- \{ 0, 1\}$ according to the official solution

parmenides51 wrote: Find the function $\displaystyle{f : \mathbb{R}-\left\{ 0\,\right\} \rightarrow \mathbb{R} }$ such that $$f(x)+f(1-\frac{1}{x}) = \frac{1}{x},\,\,\, \forall x \in \mathbb{R}- \{ 0, 1\,\}$$(-InnoXenT-) $f(x)+f(\frac{x-1}x)=\frac 1x$ $f(\frac{x-1}x)+f(\frac 1{1-x})=1-x$ $f(\frac 1{1-x})+f(x)=\frac x{x-1}$ First plus third minus second : $2f(x)=\frac 1x+x-1+\frac x{x-1}$ And so $\boxed{f(x)=\frac 12\left(\frac 1x+x-1+\frac x{x-1}\right)\quad\forall x\ne 1\text{ and }f(1)=a}$, which indeed fits, whatever is $a\in\mathbb R$