Lyub4o wrote: Find all functions $f:\mathbb R^{+} \longrightarrow \mathbb R^{+}$ so that $f(xy + f(x^y)) = x^y + xf(y)$ for all positive reals $x,y$. Let $P(x,y)$ be the assertion $f(xy+f(x^y))=x^y+xf(y)$ Let $x\ne 1$ : $P(x^{\frac 1{x-1}},x)$ $\implies$ $f(x^{\frac x{x-1}}+f(x^{\frac x{x-1}}))$ $=x^{\frac x{x-1}}+x^{\frac 1{x-1}}f(x)$ $P(x^{\frac x{x-1}},1)$ $\implies$ $f(x^{\frac x{x-1}}+f(x^{\frac x{x-1}}))=x^{\frac x{x-1}}+x^{\frac x{x-1}}f(1)$ Subtracting, we get $x^{\frac 1{x-1}}f(x)=x^{\frac x{x-1}}f(1)$ and so $f(x)=xf(1)$ $\forall x\ne 1$, still true when $x=1$ Plugging back $f(x)=ax$ in original equation, we get $a^2=1$ and so $a=1$ since $f(x)>0$ $\forall x$ Hence the unique solution : $\boxed{f(x)=x}$ $\forall x$