Nice problem!

Any ideas? I got that there must exist some point $x_0$ s.t. $f(x_0)=1$

Suppose $f$ is a non-constant function satisfying the conditions. Rewriting the beginning functional equation as \begin{align}f\left(x+\frac{f(y)}{x}\right) = f(y)f\left(\frac{y}{x}+\frac{x}{y}\right)\end{align}Claim. $f$ is injective. Proof. Let $S = \left\{c > 1 \mid \exists x > 0: f(x) = f(xc)\right\}$. Suppose $S\neq \varnothing$. Then, from $(1)$, we have $$f\left(\frac{x}{a}+\frac{a}{x}\right) = f\left(\frac{x}{b}+\frac{b}{x}\right),\forall f(a) = f(b),a < b, x > 0$$$\Rightarrow f\left(x + \frac{1}{x}\right) = f\left(cx + \frac{1}{cx}\right),\forall x > 0,c\in S$ $\Rightarrow f\left(x + \frac{1}{x}\right) = f\left(c^nx +\frac{1}{c^nx}\right),\forall x > 0,c\in S,n\in\Bbb N$. $\Rightarrow \frac{c^{2n}x^2 + 1}{c^nx} :\frac{x^2+1}{x}\in S\Rightarrow \frac{c^{2n}x^2 + 1}{c^n\left(x^2+1\right)}\in S,\forall x > 0,c\in S,n\in\mathbb N$. Observe that the function $\frac{c^{2n}x^2 + 1}{c^n\left(x^2+1\right)},x\in \mathbb R^+$ can take all values in $\left(\frac{1}{c^n},c^n\right)$. Hence, let $n\to+\infty$ we get $$c\in S,\forall c > 0$$$\Rightarrow f\left(x + \frac{1}{x}\right) = f\left(xc + \frac{1}{xc}\right),\forall x,y>0\Rightarrow f\left(x+\frac{1}{x}\right) = f\left(y+\frac{1}{y}\right),\forall 0 < x < y$ $\Rightarrow \exists C > 0: f(x) = C,\forall x > 0$ $\Rightarrow f\left(x+\frac{f(y)}{x}\right) = tf(y),\forall x,y>0$. Let $x\to+\infty$ we have $f(y) = 1,\forall y > 0$, absurd. Now, swap $x$ with $\frac{f(y)}{x}$ in $(1)$ we have $$f\left(\frac{f(y)}{x} + x\right) = f(y)f\left(\frac{xy}{f(y)} + \frac{f(y)}{xy}\right),\forall x,y>0$$By injectivity we get $$\frac{xy}{f(y)} + \frac{f(y)}{xy} = \frac{y}{x}+\frac{x}{y},\forall x,y>0$$$\Rightarrow \frac{y}{f(y)} + \frac{f(y)}{x^2y} = \frac{y}{x^2} + \frac{1}{y},\forall x,y>0$. Let $x\to+\infty$ we have $f(y) = y^2,\forall y>0$, which indeed fits.