Solve the functional equation $f(x)f(y)=f(y)f(xf(y))+\frac{1}{xy}$ Find all mappings $f$ from $\mathbb{R}_+$ to $\mathbb{R}_+$ such that for $\forall x,y \in \mathbb{R}_+$:
$$
f(x)f(y)=f(y)f(xf(y))+\frac{1}{xy}
$$
Since the domain is $\mathbb{R}_+$, we can't use the trick $x=-y$ or something like this.
Also I've tried $x = 0, 1, 2$ and found nothing useful.
What I already got is that there isn't $x$ such that $f(x) = 1$, which could be easily deduced by contradiction.

EDIT
Thanks to @Sil, we can actually use this site https://approach0.xyz to search for maths formula.
 A: Consider the statement $$P(x,y):f(x)f(y)=f(y)f(xf(y))+\frac1{xy}$$
We first look at
$$P(1,x):f(1)f(x)=f(x)f(f(x))+\frac{1}{x}$$ $$\Rightarrow f(f(x))=f(1)-\frac1{xf(x)}\ (\star)$$
To use this, we need to find another $f(f(x))$. We then look at $$P(f(x),y):f(f(x))f(y)=f(y)f(f(x)f(y))+\frac1{yf(x)}$$
$$\Rightarrow f(f(x))=f(f(x)f(y))+\frac1{yf(x)f(y)}$$
Equating them, we get $$f(1)-\frac1{xf(x)}=f(f(x)f(y))+\frac{1}{yf(x)f(y)}$$
The right-hand side is almost symmetric wrt to $x$ and $y$. Then we look at what happens if we swap $x$ and $y$
$$f(1)-\frac1{yf(y)}=f(f(x)f(y))+\frac{1}{xf(x)f(y)}$$
Subtracting them we have $$\frac1{yf(y)}-\frac1{xf(x)}=\frac{1}{f(x)f(y)}\left(\frac1y-\frac1x\right)\Rightarrow xf(x)-x=yf(y)-y$$
Since the left-hand side is just a function of $x$ and the right-hand side is just a function of $y$, they both must evaluate to a constant, say $\lambda$. Thus $$xf(x)-x=\lambda\Rightarrow f(x)=\frac{\lambda}{x}+1$$
Plugging this into $(\star)$, we get
$$\frac{\lambda x}{\lambda+x}+1=\lambda+1-\frac{1}{\lambda+x}$$
A bit of cross-multiplication, and a bit of expansion with some cancellations gives us $\lambda^2=1\Rightarrow \lambda=\pm1$. Now if $\lambda=-1$, then $f(x)=1-\frac1x$, but that is negative for $0<x<1$ and thus it is not admissible. Hence $\lambda=1$ and the only solution is $$f(x)=1+\frac1x$$
A: Based on this solution. Write the equation as
$$
P(x,y):f(x) - f(xf(y)) = \frac{1}{xyf(y)}.
$$
Then adding $P(xf(y),z)$ to $P(x,y)$ we get
$$
f(x)-f(xf(y)f(z))=\frac{1}{xyf(y)}+\frac{1}{xf(y)zf(z)}.
$$
Since LHS is symmetric in $y,z$, RHS must be too, hence
$$
\frac{1}{xyf(y)}+\frac{1}{xf(y)zf(z)}=\frac{1}{xzf(z)}+\frac{1}{xf(z)yf(y)}.
$$
Multiplying by $xyzf(y)f(z)$ and subtracting $y+z$ we find
$$
zf(z)-z=yf(y)-y.
$$
So $xf(x)-x$ must be a constant and $f(x)=\frac{c}{x}+1$. Substituting back we find $c^2=1$ and hence $c=1$ as we need only solutions in positive reals. So $f(x)=\frac{1}{x}+1$ is the only solution.
