Find all functions $f: R \rightarrow R$, such that for any real numbers $x, y$ the following is satisfied:
$$
f(x y)=f(x) f(y)-x-y .
$$
Solve: Substituting $x=y=0$ into the given equation yields $f(0)=f^{2}(0) \Rightarrow f(0)=0$ or 1. When $f(0)=0$, substituting $x=0$ gives $y=0$, which is a contradiction. When $f(0)=1$, substituting $x=1, y=-1$ gives $f(-1)=f(-1) f(1)$.
If $f(-1)=0$, then
$$
f(-x)=f(x) f(-1)-x+1=1-x \text {. }
$$

Thus, $f(x)=1+x$.
Upon verification, $f(x)=1+x$ satisfies the given conditions.
If $f(-1) \neq 0$, then
$$
f(1)=1, f(x)=f(x) f(1)-x-1,
$$

which is a contradiction.
In conclusion, $f(x)=x+1$.