The function $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfies $f(x) f(y) - f(xy) = x + y$ for all $x$, $y \in \mathbb{R}$. Find $f(x)$. The function $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfies $f(x) f(y) - f(xy) = x + y$ for all $x$, $y \in \mathbb{R}$. Find $f(x)$.
 A: With $y=1$, we can establish that
$$
f(x)f(1)-f(x)=x+1\implies f(x)=\frac{x+1}{f(1)-1}\cdot\tag{$*$}
$$
Note that the division by $f(1)-1$ is valid because $f(1)$ cannot be $1$ (if it were, $f(x)f(1)-f(x)$ would be identically $0$ whereas $x+1$ would vary with $x$.) It remains to find $f(1)$. First,
$$
x=0,y=0\implies [f(0)]^2-f(0)=0\implies f(0)=0\text{ or }f(0)=1.
$$
But $f(0)$ cannot be $0$, as we can see below:
$$
x=1,y=0\implies f(1)f(0)-f(0)=1\implies f(0)=1, f(1)=2.
$$
We conclude that $f(x)=x+1$.
A: Setting $y = 0$ gives
$f[x] f[y] - f[x y] = f[0] (f[x] - 1)$
and also
$f[x] f[y] - f[x y] = x$
$f[0] (f[x] - 1) == x$
Setting $x=0$ now gives
$f[0] (f[0] - 1) = 0$
If $f[0]$ is not zero, then $f[0] - 1$ must be zero.
So we have $f[0]=1$.
But then:
$f[x] - 1 = x$
and  
$f[x] = x + 1$
A: Setting $y = 0$, we get
[f(0) f(x) - f(0) = x,]
so $f(0) [f(x) - 1] = x$ for all $x$. In addition, setting $x = 0$, we get $f(0) [f(0) - 1] = 0$, so $f(0) = 0$ or $f(0) = 1$. But $f(0) [f(x) - 1] = x$ for all $x$, so $f(0)$ cannot be 0, which means $f(0) = 1$. Then $f(x) - 1 = x$, or $f(x) = \boxed{x + 1}$ for all $x$. It is easy to verify that this solution works.
