Functional equation $(x + y)f(f(x)y) = x^2 f(f(x) + f(y))$

Find all functions $f:\mathbb{R}^+ \to \mathbb{R}^+$ such that $$(x + y)f(f(x)y) = x^2 f(f(x) + f(y)) \mbox{, for all } x,y\in\mathbb{R}^+.$$ I tried out various substitutions such as $x=y$, $x=1$, $x+y=x^2$, and nothin' works.

• Can you think of any such functions? – Pat Devlin Dec 29 '16 at 14:44
• Trying $y = 1/f(x)$ looks interesting. – Pat Devlin Dec 29 '16 at 14:48
• I didn't found any of them. You are right, pair $[x=1, y = 1/f(1)]$ can be useful. – Pathbreaker Dec 29 '16 at 15:16
• Are there any points you know? – Pat Devlin Dec 29 '16 at 15:41
• Or try $y = x / f(x)$ – Pat Devlin Dec 29 '16 at 15:46

$$\frac{f(f(x)y)}{x^2} = \frac{f(f(x) + f(y))}{x + y }\tag{1}$$ Using the symmetry of the right side of $(1)$: $$\frac{f(f(x)y)}{x^2} = \frac{f(f(x) + f(y))}{x + y }= \frac{f(f(y) + f(x))}{y + x }=\frac{f(f(y)x)}{y^2} \implies\\$$ $$\frac{f(f(x)y)}{x^2} =\frac{f(f(y)x)}{y^2} \tag{2}$$ With $x=y$ in $(1)$ we have : $$\frac{f(xf(x))}{x} = \frac{f(2f(x))}{2} \tag{3}\\$$ One solution to $(2)$ is : $f(x)=x^2$. A solution to $(3)$ is: $f(x) = x$.
Lemma 1 : if $f(x)$ exists it must be injective.
Let $f(a)=f(b)$: $$A: \enspace \frac{f(af(b))}{a}=\frac{f(af(a))}{a} = (3) = \frac{f(2f(a))}{2} = \frac{f(2f(b))}{2} = (3) = \frac{f(bf(b))}{b}=\frac{f(bf(a))}{b}\\ B: \enspace (2) \implies \frac{f(af(b))}{b^2} = \frac{f(bf(a))}{a^2} \\ A \enspace and \enspace B: \enspace \implies \frac{a}{b} \cdot \frac{f(bf(a))}{b^2} = \frac{f(af(b))}{b^2} = \frac{f(bf(a))}{a^2} \implies a^3=b^3 \implies a = b \enspace \square \\$$
Now we prove $(1)$ has no solutions because $f(x) <0$ for some $x$ .
Let: $x > 1$. $$(1) \implies \frac{f(f(x)(x^2-x))}{x^2} = \frac{f(f(x) + f(x^2-x))}{x+ x^2 -x}\\ f(f(x)(x^2-x)) = f(f(x) + f(x^2-x)) \implies \text{ with injectivity Lemma 1 } \implies \\ f(x)(x^2-x) = f(x) + f(x^2-x) \implies x^2-x=\frac{f(x)}{f(x)} + \frac{ f(x^2-x)}{f(x)} \implies \\ x^2-x -1 = \frac{f(x^2-x)}{ f(x)}$$ We see that $x^2-x -1 < 0$ for $1 <x < \frac{1+\sqrt{5}}{2}$ ( $\implies f(x^2-x) < 0 \lor f(x) < 0$ ) and conclude $(1)$ has no solutions in the range $f:\mathbb{R}^+ \to \mathbb{R}^+$ $\enspace \square$
(Note that : $\frac{1+\sqrt{5}}{2}$ is the famous golden ratio ).