Functional equation : $f(xf(y))=yf(x)$

Find all function $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ satisfying

(i) $f(xf(y))=yf(x)$ for all positive real numbers, $x, y$

(ii) $f(x)$ is bounded function for domain interval $(1,\infty)$

Please help me check my work below. Thank you.

Let $P(x,y)$ denote $f(xf(y))=yf(x),\;\;\forall x, y \in \mathbb{R^+}$.

Let $y_1, y_2 \in \mathbb{R}^+$ such that $f(y_1)=f(y_2)$.

Consider $P(x,y_1)$ and $P(x,y_2)$ : we have

$f(xf(y_1))=f(xf(y_2))= y_1f(x) = y_2f(x)$ then $y_1=y_2$ so $f$ is injective.

Consider $P(1,1)$ : $f(f(1))=f(1) \rightarrow f(1)=1$

substitute $x=1$ in (i), we have $f(f(y))=y$.

Consider $P(x, x)$ : $f(xf(x))=xf(x)$

Let $xf(x)=z$ so $f(z)=z, \;\exists z \in \mathbb{R}^+$

If $z>1$ then $P(x,f(y)) : f(x)f(y)=f(xy)$ so $f(z)f\left(\frac{1}{z}\right)=f(1)=1$ so $f\left(\frac{1}{z}\right) = \frac{1}{z}$

then $f(z^2)=f(z)f(z)=z^2$ , $f(z^4)=f(z^2)f(z^2)=z^4$ so $f(z^{2^n})= z^{2^n}, \;\forall n \in \mathbb{N}$, contradict (ii)

If $z<1$ then $\frac{1}{z}>1$. Since $f\left(\frac{1}{z}\right) = \frac{1}{z}$ so $f\left(\frac{1}{z^2}\right) = f\left(\frac{1}{z}\right)f\left(\frac{1}{z}\right)=\frac{1}{z^2}$

then $f\left(\frac{1}{z^{2^n}}\right)=\frac{1}{z^{2^n}}$, contradict (ii)

Thus $xf(x)=1$, $f(x)=\frac{1}{x} ,\;\;\forall x \in \mathbb{R^+}$

Answer check :

(i) $f(xf(y))=\frac{y}{x}=yf(x)$

(ii) $f(x)=\frac{1}{x} < f(1)=1 ,\;\;\forall x \in (1,\infty) \;\;\blacksquare$