This problem was proposed by me and Blastoor.

very nice question.

We claim the only solution is $\boxed{f(x) = x+1}$. This works. Let $P(x,y)$ denote the given assertion. $P(0,1): f(f(1)) = 3$. So also $f(f(0) + 1) = 3$. $P(1,0): f(-1) = 0$. $P(x,1): f(0) + f(x + f(1)) = f(x) + 3$. $P(0,x): f(0) + f(f(x)) = xf(0) +3$. Setting $x = -1$ here gives $f(0) = 1$, so $f(f(x)) = x+2$ for all $x$. Thus, $f$ is bijective. $P(x,1): f(x+2) = f(x) + 2$. $P(x,2): f(x+3) = f(x) + 3 $. Now, \[ f(x) + 3f(x+3) = f((x+1) + 2) = f(x+1) + 2,\]which implies $f(x) + 1 = f(x+1)$. $P(1,x): f(x-1) + f(f(x) + 1) = 2x + 3$. Since $f(f(x) + 1) = f(f(x)) + 1 = x+3$, we get $f(x-1) = x$, so $f(x) = x+1$ for all $x$, as desired.

Here is my solution: https://calimath.org/pdf/BalticWay2022-5.pdf And I uploaded the solution with motivation to: https://youtu.be/Gp8U-vCuMTY

Let $P(x,y)$ denote the assertion $f(xy-x)+f(x+f(y)) = yf(x)+3$. Note that $P(1,-1)$ gives $f(0)+f(-1+f(1))=f(0)+f(f(0)) =f(-1)+3$ and that $P(0,0)$ gives $f(0)+f(f(0)) = 3$. Thus, $f(-1) =0$. Also note that $P(0,-1): f(0)+f(0) = -f(0) +3$. Thus, $f(0) = 1$ and $f(1) = 2$. We see, \[P(x,1): 1+f(x+2)=f(x)+3 \implies f(x+2) = f(x)+2\]and \[P(x,2): f(x)+f(x+3) = 2f(x)+3 \implies f(x+3)=f(x)+3\]. Combining the two equations give $f(x+1) = f(x) + 1$. We can further note that \[P(0,x): 1+f(f(y)) = x+3\]and again note that \[P(1,x): f(x-1)+f(1+f(x)) = 2x+3 \implies f(x-1)+x+3 = 2x+3 \implies \boxed{f(x-1) = x}\] oops sorry for (accidentally) plagiarizing u arnev, also thanks for helping me realize that $f(x+2) = f(x)+2$ doesn't finish the problem

Here is a way to obtain the values of $f(0)$ and $f(1)$ without assuming that $f(1)-f(0)=1$.

Solved with @enarmonia

We claim that the unique solution is $f(x) \equiv x + 1$, which evidently works. Let $A(x, y)$ denote the given assertion. $A(0, 1)$ gives us $f(f(1)) = 3$. Using this and $f(1) = f(0) + 1$, $A(1, 0)$ gives us $f(-1) = 0$. Then, $A(-1, -1)$ gives us $f(2) = -2f(-1) + 3 = 3$. Then, $A(x, 2)$ is \[ f(x) + f(x + 3) = 2f(x) + 3 \implies f(x + 3) = f(x) + 3. \qquad (1)\]Using this, we get that $f(5) = f(2) + 3 = 6$. Then, $A(1, 5)$ gives us \[ (f(1) + 3) + (f(1) + 6) = f(4) + f(7) = 5f(1) + 3 \implies f(1) = 2. \]In addition, $f(0) = 1$ follows. So, $A(0, x)$ gives us \[ f(0) + f(f(x)) = xf(0) + 3 \implies f(f(x)) = x + 2. \qquad (2)\]Tripling the composition in (2) gives us $f(x + 2) = f(x) + 2$. Joining this with (1), we get that \[ f(x + 1) \overset{(1)}{=} f(x - 2) + 3 \overset{(2)}{=} f(x) + 1. \qquad (3)\]Finally, using (2) and (3), $A(1, x)$ gives us \[ f(x - 1) + f(f(x)) + 1 = 2x + 3 \implies f(x - 1) = x. \]So, no solutions exist other than $f(x) \equiv x + 1$.

We claim the only solution is $\boxed{f(x)\equiv x+1}$. It is easy to check that this works. Let $P(x,y)$ denote the given assertion. We have \begin{align*} P(0,x)&\implies f(0)+f(f(x))=xf(0)+3&&\implies f(f(1))=3\\ P(1,0)&\implies f(-1)+f(1+f(0))=3&&\implies f(-1)=0\\ P(0,-1)&\implies f(0)+f(0)=-f(0)+3&&\implies f(0)=1 \end{align*}so $f(f(x))=x+2$. Thus $f(1)=2$ so $f(2)=f(f(1))=3$. We have \begin{align*} P(x,1)&\implies f(0)+f(x+2)=f(x)+3&&\implies f(x+2)=f(x)+2\\ P(x,2)&\implies f(x)+f(x+3)=2f(x)+3&&\implies f(x+3)=f(x)+3 \end{align*}so $f(x+1)=f(x)+1$ for all $x$. Then $P(1,x+1)$ gives $f(x)+f(1+f(x+1))=2(x+1)+3$. Since $f(1+f(x+1))=2+f(f(x))=x+4$, we have $f(x)=x+1$, as desired. $\square$

Solution without using the given equality. $P(x,2) \implies f(x+f(2)) = f(x)+3$. Let $ a = f(2), b = 3$ where $a \neq 0$. Then $$P(x+a,y) - P(x,y) \implies f( (x+a) (y-1) ) - f(x(y-1)) + b = by \overset{y \rightarrow y+1} \implies $$$$f(xy+ay) - f(xy) = by \overset{x \rightarrow x/y, y \neq 0} \implies f(x+ay) = f(x) + by \overset{y \rightarrow \frac{y}{a}} \implies $$$$f(x+y) = f(x)+\frac{b}{a}y \implies f(x)+\frac{b}{a}y = f(y)+\frac{b}{a}x \implies f(x) - \frac{b}{a}x = c$$The only linear solution found by plugging is $f(x) \equiv x+1$