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.
Thanks in advance for your reply.
 A: $$\frac{f(f(x)y)}{x^2} =  \frac{f(f(x) + f(y))}{x + y }\tag{1}$$
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 ).
