jmc

We write the functional equation as $$f(x)f(y)-f(xy)=3x+3y+6.$$Setting $x=y=0,$ we get $$f(0{)}^2-f(0)=6.$$Then $f(0{)}^2-f(0)-6=0,$ which factors as $(f(0)-3)(f(0)+2)=0.$ Hence, $f(0)=3$ or $f(0)=-2.$

Setting $y=0,$ we get $$f(0)f(x)-f(0)=3x+6.$$Then $$f(x)-1=\frac{3x+6}{f(0)},$$so $$f(x)=\frac{3x+6}{f(0)}+1.$$If $f(0)=3,$ then $f(x)=x+3,$ which does satisfy the functional equation. If $f(0)=-2,$ then $$f(x)=-\frac{3}{2}x-2,$$which does not satisfy the functional equation. Therefore, $f(x)=\boxed{x+3}.$