parmenides51 wrote: Let ${R^* = R-{0}}$ be the set of non-zero real numbers. Find all functions ${f : R^* \rightarrow R^*}$ satisfying ${f(x) + f(y) = f(xy f(x + y))}$, for ${x, y \in R^*}$ and ${ x + y\ne 0 }$. Let $P(x,y)$ be the assertion $f(x)+f(y)=f(xyf(x+y))$ If $f(x)\ne \frac 1x$ for some $x\ne 0$, then : $P(\frac 1{f(x)},x-\frac 1{f(x)})$ $\implies$ $f(\frac 1{f(x)})=0$, impossible. So $\boxed{f(x)=\frac 1x\quad\forall x\ne 0}$ which indeed is a solution.