functional equation problem in competition: $ f ( x y ) = f ( x ) f ( y ) - f ( x + y ) + 1 $ 
Find all $ f : \mathbb Q \to \mathbb Q $ such that $ f ( 1 ) = 2 $ and
$$ f ( x y ) = f ( x ) f ( y ) - f ( x + y ) + 1 $$
for all $ x , y \in \mathbb Q $.

thank you very much!

Put $y=1$. Than we have 
$$f(x)=f(x)\cdot f(1)- f(x+1)+1$$
Hence 
$$f(x)=2f(x)- f(x+1)+1$$
Hence
$$f(x+1)=f(x)+1$$.
So we have $f(0)=1$.
Now we say $x=-y$, we have 
$$f(-x^2)=f(x)f(-x)$$
When we have
$$0=f(-1)=f(-2\cdot \frac{1}{2}) = f(-2) f(\frac{1}{2})- f(-\frac{3}{2}) +1$$ 
So we have
$$0= -1\cdot f(\frac{1}{2}) - f(\frac{1}{2}) +3$$
Hence 
$$0= -1(2\cdot f(\frac{1}{2})-3) $$
Hence $f(\frac{1}{2})=1$
You need more help? 
In general we have for $x\in \mathbb{Z}$ 
\begin{align*}
0&=f(-1)\\
&=f(-x\cdot \frac{1}{x}) \\
&=f(-x) \cdot f(\frac{1}{x}) - f(-x+\frac{1}{x})+1\\
&= (-x+1)\cdot f( \frac{1}{x}) -  f(\frac{1}{x}) +x\\
&= -x \cdot( f(\frac{1}{x}) -1)
\end{align*}
