Find all functions $f$ defined in the set of Real Numbers without zero, satysfying equation $$f(xyf(x+y))=f(x)+f(y)$$ For all $x\neq 0, y\neq 0$ and $x+y\neq0$ Thanks
Edit: I found out that function $\frac{1}{x}$ is a solution but I dont know how to prove there are no others.
Suppose that $f(u)\neq u^{-1}$ for some $u\in\Bbb{R}^{\times}$, so that $x:=u-f(u)^{-1}\in\Bbb{R}^{\times}$. Let $y:=f(u)^{-1}\in\Bbb{R}^{\times}$ so that $x+y=u\in\Bbb{R}^{\times}$ and $f(x+y)=f(u)=y^{-1}$. Then $$f(x)+f(y)=f(xyf(x+y))=f(x),$$ and hence $f(y)=0$, a contradiction. So $f(u)=u^{-1}$ for all $u\in\Bbb{R}^{\times}$.