Let function $f:R\setminus 0\to R$ such
(1): $$\dfrac{f(x)}{x}=f\left(\dfrac{1}{x}\right),\forall x\neq 0$$ (2): for any $x,y$ such $$f(x)+f(y)=f(x+y)+1,\forall x+y\neq 0$$
Find $f$
Let $P(x,y)$ be the assertion $f(x)+f(y)=f(x+y)+1$, $Q(x)$ be the assertion $\dfrac{f(x)}{x}=f\left(\dfrac{1}{x}\right)$. Subtracting $Q(-1)\Longrightarrow f(-1)=0$ $$P(1,0)\Longrightarrow f(0)=1$$ Subtracting $P(-1,-1)\Longrightarrow f(-2)=-1$ Subtracting $P(-1,-2)\Longrightarrow f(-3)=-2$.
But I can't prove $f(x)=x+1$
Letting $g(x)=f(x)-1$ and aditionaly $g(0)=0$, (2) is equivalent to $g(x)+g(y)=g(x+y)$. From this we can deduce, among other things, $g(2x)=2g(x)$ and $g(\frac{x}{2})\frac{g(x)}{2}$.
Now (1) is saying that for $x\neq 0$ $\frac{g(x)}{x}+\frac{1}{x}=g(\frac{1}{x})+1$. By replacing $x$ with $2x$ we get $\frac{g(2x)}{2x}+\frac{1}{2x}=g(\frac{1}{2x})+1$, so we have, by above, $\frac{g(x)}{x}+\frac{1}{2x}=\frac{g(\frac{1}{x})}{2}+1$. Multiplying the latter equation by $2$ and subtracting the first one gives $\frac{g(x)}{x}=1,g(x)=x$.
Thanks Hagen for pointing out a (minor) flaw in earlier argument.