The answer is $f\equiv x$ .This obviously satisfies the condition . It is easy to see that $f$ is injective .And surjective at $(1+\frac{1}{f(1)},\infty)$ On RHS take $f(y)$ very big and $x$ very small so $f$ is surjective at $(1,\infty)$ Take $x\rightarrow \frac{1}{f(x)}$ so RHS is symmetric so $f(\frac{1}{f(x)})-\frac{x+1}{f(x)}$ is constant. Let say its equal to $c$ $f(f(x)+f(\frac{1}{f(y)})-c)=\frac{1}{f(y)}+x+1$ . We could replace $\frac{1}{f(y)}$ with $y<1$ $f(f(x)+f(y)-c)=x+y+1$ So if $x+y=z+t$ and $y,t<1$ then $f(x)+f(y)=f(z)+f(t)$ then we can get (with some tricks ) if $x+y=z+t$ then $f(x)+f(y)=f(z)+f(t)$ so $f(x)+f(y)=2f(\frac{x+y}{2})$ for all $x, y \in R^+$ Let $g(x)=f(x+1)-f(1)$ such that $g:(-1,\infty) \rightarrow (-f(1),\infty)$ then $g(\frac{x}{2})=\frac{g(x)}{2}$ for $x \in (-1 ,\infty)$ and $g(u)+g(v)=g(u+v)$ for $ u,v \in (-1,\infty)$ and $u+v>-1$ Obviously $g(0)=0$ and $g$ is odd function . Lets extend the function $g$ to $h$ such that if $x \in R_0^+$ $h(x)=g(x)$ else $h(x)=-g(-x)$ $h:R\rightarrow R$ obviously $h$ is additive and for $x \in (-1,\infty)$ $h(x) \in (-f(1),\infty) $ so $h$ has a bound for non-trivial interval this means $h(x)=kx$ means $g(x)=kx $ so $f$ is a lineer function if we check $f(x)\equiv x$ $\blacksquare$

VicKmath7 wrote: Find all functions $f: \mathbb{R^{+}} \rightarrow \mathbb {R^{+}}$ such that $f(f(x)+\frac{y+1}{f(y)})=\frac{1}{f(y)}+x+1$ for all $x, y>0$. Proposed by Dominik Burek, Poland See this as $f(f(x)+g(y))=x+h(y)$ then by a well know lemma in $R^+$ we get that: $g(y)-h(y)=c\Rightarrow \frac{y+1}{f(y)}-1-\frac{1}{f(y)}=c\Rightarrow f(y)=\frac{y}{c+1}$ And esily we get $f(x)=x$ https://artofproblemsolving.com/community/q3h2854866p25319580