Find all function $f$: $\Bbb R^+ \rightarrow \Bbb R^+$ such that $f\left(\frac{f(x)}{y}\right) = yf(y)*f(f(x))$

Question

Find all function $f$: $\Bbb R^+ \rightarrow \Bbb R^+$ such that $$f\left(\frac{f(x)}{y}\right) = yf(y)*f(f(x))$$ for all $x, y\in\Bbb R^+$

My attempt:-

Put $y = 1$,
$f(f(x)) = f(1)*f(f(x))$ $\implies f(1) = 1 \hspace{1cm} \{\because f(x) > 0\}$

Put $y = f(x)$,
$f(1) = f(x)*f(f(x))^2$
$\implies f(f(x)) = \cfrac{1}{\sqrt{f(x)}}$ $\implies f(y) = \cfrac{1}{\sqrt{y}}$ and finished.
But when I saw the solution it says "We cannot conclude from it". Is this because I assume $y = f(x)$ and I need to do more steps? Or something else?

EDIT:-
There are so many confusions in comments. This is the solution of the question. I didn't understand from last $3^{rd}$ line of LHS

Edit again:- After so many confusion what I understood is, I proved $f(y) = \cfrac{1}{\sqrt{y}}$ for $y \in$ {all possible values of $f(x)$ for $x \in \Bbb R^+\}$ but I have to prove for $y \in \Bbb R^+$ ($x$ and $y \in \Bbb R^+$ is given but not $f(x) \in \Bbb R^+$). Since we don't know yet if $f(x) \in \Bbb R^+$, the way we prove $f(x) \in \Bbb R^+$ is using $f(\frac{u}{v})$ where $\frac{u}{v} = y\in \Bbb R^+$ as stated by Anne in the answer, Thanks Anne for helping me in understanding the solution.

Answer 1

Thank you for having added the solution. Let us read it together, comparing it with your attempt.

You proved that $f(1)=1$ (so do they), and (instantiating the hypothesis with $y=f(x)$) $$\forall y\in{\rm im}(f),\quad f(y)=\frac1{\sqrt y},$$ and you could not conclude that the same holds for every $y\in\Bbb R^+,$ because you did not know yet that $f$ is onto.

What they proved instead (instantiating with $y\in{\rm im}(f)$ but not related to $x$) is: $$\forall u,v\in{\rm im}(f),\quad f(u/v)=\frac1{\sqrt{u/v}}.$$ In order to conclude that $$\forall y\in\Bbb R^+,\quad f(y)=\frac1{\sqrt y},$$ they managed to prove that every $y\in\Bbb R^+$ can be written $\frac uv$ for some $u,v\in{\rm im}(f),$ by the following trick (which I simplify without losing its efficiency, instantiating the hypothesis with $x=1$ and using thrice the fact that $f(1)=1$): $$y=\frac{f(1/y)}{f(y)}.$$ This ends the proof (the converse fact that $y\mapsto\frac1{\sqrt y}$ is a solution goes without saying).