Outline from Contest: Take $y = x$, $y=0$ and $x=1$ and conclude that $f(x) = \frac{2}{x+1}-1$ - boring problem. Remark: Problems are ordered by difficulty (mod 4), only $5$ of the problems were released due to confidentiality. @below I meant do these three things separately but yes rules are indeed made to be broken.

L567 wrote: satisfying $x \neq 0$ and $y \neq x$ bora_olmez wrote: Take $y = x$, $x=0$ Rules are made to be broken

Shouldn't it be $y\ne -x$ because you can set $y=-x$ and that gives a contradiction?

L567 wrote: Let $D$ be the set of real numbers excluding $-1$. Find all functions $f: D \to D$ such that for all $x,y \in D$ satisfying $x \neq 0$ and $y \neq x$, the equality $$(f(f(x))+y)f \left(\frac{y}{x} \right)+f(f(y))=x$$holds. I suppose you missed the constraint $y\ne -x$ in the domain of functional equation. Else, there are obviously no solution. If so : Let $P(x,y)$ be the assertion $(f(f(x))+y)f(\frac yx)+f(f(y))=x$, true for all $x\ne 0$ and $y\notin\{-1,-x,x\}$ Let $a=f(0)$ ($a\ne -1$) and $c=f(1)$ $P(x,0)$ $\implies$ $af(f(x))=x-f(a)$ and so $a\ne 0$ and $f(f(x))=\frac xa-\frac{f(a)}a$ $\forall x\ne 0$ Plugging this in $P(x,y)$, we get : New assertion $Q(x,y)$ : $(\frac xa-\frac{f(a)}a+y)f(\frac yx)+\frac ya-\frac{f(a)}a=x$, true for all $x\ne 0$ and $y\notin\{-1,0,-x,x\}$ $Q(x,xy)$ $\implies$ $x((1+ay)f(y)+y-a)=f(a)+f(a)f(y)$, true for all $x\ne 0$ and $y\notin\{-\frac 1x,-1,0,+1\}$ We know that $a\ne 0,-1$. If $a\ne 1$, choosing here above $x\ne a$ and $y=-\frac 1a$, we get $x(-\frac 1a-a)=f(a)+f(a)f(-\frac 1a)$ $\forall x\ne 0,-1,a$, impossible. So $a=1$ and above result is $x((1+y)f(y)+y-1)=c(1+f(y))$ for all $x\ne 0$ and $y\notin\{-\frac 1x,-1,0,+1\}$ And so $(1+y)f(y)+y-1=c(1+f(y))=0$ $\forall y\notin\{-1,0,1\}$ And so $c=0$ and $f(y)=\frac{1-y}{1+y}$ $\forall y\notin\{-1,0,1\}$, still true when $y=0$ or $y=1$ And so $\boxed{f(x)=\frac{1-x}{1+x}\quad\forall x\ne -1}$, which indeed fits

L567 wrote: Let $D$ be the set of real numbers excluding $-1$. Find all functions $f: D \to D$ such that for all $x,y \in D$ satisfying $x \neq 0$ and $y \neq -x$, the equality $$(f(f(x))+y)f \left(\frac{y}{x} \right)+f(f(y))=x$$holds. $\color{blue}\boxed{\textbf{Answer:}f\equiv \frac{1-x}{x+1}}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $$(f(f(x))+y)f \left(\frac{y}{x} \right)+f(f(y))=x...(\alpha)$$In $(\alpha) x=y\neq 0:$ $$\Rightarrow (f(f(x))+x)f(1)+f(f(x))=x$$$$\Rightarrow f(f(x))(f(1)+1)=x(1-f(1))$$Since $f(1)\neq -1:$ $$\Rightarrow f(f(x))=x(\frac{1-f(1)}{f(1)+1})$$$$\Rightarrow f(f(x))=cx, \forall x\in D, x\neq 0...(I)$$Replacing $(I)$ in $(\alpha):$ $$\Rightarrow (cx+y)f(\frac{y}{x})+cy=x, \forall x\in D, x\neq 0...(\beta)$$In $(\beta) x=1, y\neq -1:$ $$\Rightarrow (c+y)f(y)+cy=1$$$y\neq -c:$ $$\Rightarrow f(y)=\frac{1-cy}{c+y}, \forall y\in D, y\neq 0,-1,-c...(II)$$Since $\frac{1-cy}{c+y}\neq -1, \forall y\in D, y\neq 0,-1,-c$ $$\Rightarrow 1-cy\neq -c-y, \forall y\in D, y\neq 0,-1,-c$$$\color{red}\boxed{\textbf{If c}\neq \textbf{1}}$ $\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow y\neq \frac{c+1}{c-1}, \forall y\in D, y\neq 0,-1,-c$$$$\Rightarrow \frac{c+1}{c-1}=0,-1,-c $$$$\Rightarrow c\in \{ -1, 0\}$$$\color{red}\boxed{\textbf{If c}=\textbf{-1}}$ By $(II):$ $$\Rightarrow f(x)=\frac{x+1}{x-1}, \forall y\in D, y\neq 0,-1,1$$Replacing in $(I):$ $$\Rightarrow \frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1}=-x, \forall y\in D, y\neq 0,-1,1$$$$\Rightarrow x=-x, \forall y\in D, y\neq 0,-1,1(\Rightarrow \Leftarrow)$$$\color{red}\boxed{\textbf{If c}=\textbf{0}}$ By $(II):$ $$\Rightarrow f(x)=\frac{1}{x}, \forall y\in D, y\neq 0,-1$$Replacing in $(I):$ $$\Rightarrow \frac{1}{\frac{1}{x}}=0, \forall y\in D, y\neq 0,-1$$$$\Rightarrow x=0, \forall y\in D, y\neq 0,-1(\Rightarrow \Leftarrow)$$$\color{red}\rule{24cm}{0.3pt}$ $\color{red}\boxed{\textbf{If c}=\textbf{1}}$ $\color{red}\rule{24cm}{0.3pt}$ By $(II):$ $$\Rightarrow f(x)=\frac{1-x}{x+1}, \forall y\in D, y\neq 0,-1$$It is easy to verify that it complies $\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow \boxed{f(x)\equiv \frac{1-x}{x+1}\textbf{is the only solution}_\blacksquare}$$$\color{blue}\rule{24cm}{0.3pt}$