Amin12 wrote:

Find all functions $f:\mathbb{R^+}\rightarrow\mathbb{R^+}$ such that
$\frac{x+f(y)}{xf(y)}=f\left(\frac{1}{y}+f\left(\frac{1}{x}\right)\right)$ for all positive real numbers $x$ and $y$ .

Equivalently, $\frac{1}{x}+\frac{1}{f(y)}=f\left(\frac{1}{y}+f\left(\frac{1}{x}\right)\right)$ for all $x,y>0$ . Put $t=\tfrac{1}{x}$ thus $t+\tfrac{1}{f(y)}=f\left(f(t)+\tfrac{1}{y}\right)$ for all $t,y>0$ . Observe that as $t \rightarrow \infty$ we see $\mathbb{R}(f)$ has no upper bound. Thus, for any $\varepsilon>0$ it is possible to pick $y$ with $\tfrac{1}{f(y)}<\varepsilon$ ; hence $\cup [\tfrac{1}{f(y)}, \infty)$ is a subset of $\mathbb{R}(f)$ , consequently $f$ is surjective over $\mathbb{R}^{+}$ . If $f(a)=f(b)$ then plugging $t=a$ and $t=b$ subsequently, we conclude $a=b$ or $f$ is injective. Thus, $f$ is a bijection on positive reals.

Now substitute $y=\tfrac{1}{f(z)}$ yielding $f(f(t)+f(z))=t+\frac{1}{f\left(\frac{1}{f(z)}\right)}$ for all $t,z>0$ . Swapping $t,z$ fixes the LHS, hence $\frac{1}{f\left(\frac{1}{f(z)}\right)}=z+C$ for some constant $C$ and all $z>0$ . Hence $f(f(t)+f(z))=t+z+C$ for all $t,z>0$ . Playing the Devil's trick again, we put $x=f(z)$ in the original equation; so $\frac{1}{f(z)}+\frac{1}{f(y)}=f\left(\frac{1}{y}+\frac{1}{z+C}\right)$ and swap $y,z$ ; injectivity of $f$ then yields that $y \mapsto \frac{1}{y}-\frac{1}{y+C}$ is a constant function. Thus, $C=0$ and so $f(f(y)+f(z))=y+z$ for all $y,z>0$ . Now plug $x \mapsto f\left(\frac{1}{x}\right)$ in $f\left(\frac{1}{f(x)}\right)=\frac{1}{x}$ to conclude that $f\left(\frac{1}{x}\right)=\frac{1}{f(x)}$ for all $x>0$ . Now we immediately get $f(f(x))=x$ for all $x>0$ and so $f(a+b)=f(a)+f(b)$ (putting $a=f(y), b=f(z)$ ); hence $f$ is additive too. Now $f$ is strictly increasing so if $f(x_0)>x_0$ then $x_0=f(f(x_0)>x_0$ and vice-versa. Thus, $f(x_0)=x_0$ for all $x_0>0$ and $f$ is the identity function. It also works. $\blacksquare$