How can I solve this functional equations

Question

Find $f: \Bbb R \to \Bbb R$ such that $f(x) = xf\left(\frac1x\right) = 1+f(x+y)-f(y)$, where $x,y \in \Bbb R \setminus\{0\}$ are arbitrary.

Answer 1

We use the following relations: one directly given, $$f(x)=x\,f\left(\frac1x\right)\tag1,$$ the other one derived by replacing $x$ and $y$ by $x/2$: $$f(x)=2\,f\left(\frac{x}2\right)-1\tag2.$$ Now, we evaluate the same expression $f(2/x),$ in two different ways. First by applying (2) with $x\rightarrow2/x$, followed by (1) with $x\rightarrow1/x$: $$f\left(\frac2x\right)=2f\left(\frac1x\right)-1=\frac{2f(x)}x-1$$ Second by applying (1) with $x\rightarrow2/x$, followed by (2): $$f\left(\frac2x\right)=\frac2x\,f\left(\frac{x}2\right)=\frac{2\,f\left(\frac{x}2\right)}x=\frac{f(x)+1}x.$$ Comparing both results, we obtain $$f(x)+1=2f(x)-x$$ and finally $$f(x)=1+x.$$ Obviously, it satisfies the given equations.