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.

$$\frac{f(f(x)y)}{x^2} =  \frac{f(f(x) + f(y))}{x + y }\tag{1}$$
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) $:
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