If $f$ is not constant, $f(0)=-1$.Otherwise, setting $y=0$, we get $f(x)=1$. Set $y=1$ to have $f(x)=f(x)f(1)+f(x+1)-1$. Now, use this in induction to prove that $f(n)=n-1$ for $n\in\mathbb N_0$. We have $f(x)=f(x+1)-1$. Set $x=-1,y=1$ and we have $f(-1)=-2$. Then use induction again to prove that $f(x)=x-1$ for $x\in\mathbb Z$. Since $f(x+1)=f(x)+1$, we have $f\left(\frac uv+n\right)=f\left(\frac uv\right)+n$ for $n\in\mathbb Z$, so $f\left(\frac1n+m\right)=f\left(\frac1n\right)+m$. Again, setting $x=n,y=\frac1n$ gives $f(1)=f(n)f\left(\frac1n\right)+f\left(n+\frac1n\right)-1$ or $(n-1)f\left(\frac1n\right)+n+f\left(\frac1n\right)=1$ inferring $f\left(\frac1n\right)=\frac1n-1$. Set $x=m,y=\frac1n$ to get \[f\left(\frac mn\right)=f(m)f\left(\frac1n\right)+f\left(m+\frac1n\right)-1=(m-1)\left(\frac1n-1\right)+m+\frac1n-1-1=\frac mn-1\]

utkarshgupta wrote: Find all functions $f: \mathbb{Q} \to \mathbb{R}$ such that $f(xy)=f(x)f(y)+f(x+y)-1$ for all rationals $x,y$ The only constant solutions are $\boxed{\text{S1 : }f(x)=1\text{ }\forall x\in\mathbb Q}$ and $\boxed{\text{S2 : }f(x)=-1\text{ }\forall x\in\mathbb Q}$. So let us from now look only for non constant solutions. Let $P(x,y)$ be the assertion $f(xy)=f(x)f(y)+f(x+y)-1$ Let $a=f(1)$ If $f(0)\ne -1$, then $P(x,0)$ $\implies$ $f(x)=1$, impossible since non constant. So $f(0)=-1$ $P(1,1)$ $\implies$ $f(2)=-a^2+a+1$ $P(2,2)$ $\implies$ $(a^2-a-1)^2=1$ and so $a\in\{-1,0,1,2\}$ 1) If $a=-1$ $P(x,1)$ $\implies$ $f(x+1)=2f(x)+1$ and so $f(x+n)=2^nf(x)+2^n-1$ and $f(n)=-1$ Then $P(x,n)$ implies $f(nx)=(2^n-1)(f(x)+1)-1$ Setting there $x=\frac mn$, we get $f(\frac mn)=-1$ and $f(x)$ constant, impossible there. 2) If $a=0$ $P(x,1)$ $\implies$ $f(x+1)=f(x)+1$ and so $f(x+n)=f(x)+n$ and $f(n)=n-1$ Then $P(x,n)$ implies $f(nx)=nf(x)+n-1$ Setting there $x=\frac mn$, we get $f(\frac mn)=\frac mn-1$ and so $\boxed{\text{S3 : }f(x)=x-1\text{ }\forall x\in\mathbb Q}$ which indeed is a solution. 3) If $a=1$, then $P(-1,1)$ gives a contradiction. 4) If $a=2$ : $P(x,1)$ $\implies$ $f(x+1)=1-f(x)$ and so $f(x+2)=f(x)$ and $f(2)=-1$ Then $P(\frac 12,2)$ implies contradiction