Find all functions \( f: \mathbb{R} \rightarrow \mathbb{R} \) that satisfy the conditions 
\[ f(1+x y)-f(x+y)=f(x) f(y) \quad \text{for all } x, y \in \mathbb{R} \]
and \( f(-1) \neq 0 \).
The only solution is the function $f(x)=x-1, x \in \mathbb{R}$. We set $g(x)=f(x)+1$ and show that $g(x)=x$ for all real $x$. The conditions take the form 
$$ g(1+x y)-g(x+y)=(g(x)-1)(g(y)-1) \quad \text { for all } x, y \in \mathbb{R} \text { and } g(-1) \neq 1 $$ 
Denote $C=g(-1)-1 \neq 0$. Setting $y=-1$ in (1) gives 
$$ g(1-x)-g(x-1)=C(g(x)-1) . $$ 
Set $x=1$ in (2) to obtain $C(g(1)-1)=0$. Hence $g(1)=1$ as $C \neq 0$. Now plugging in $x=0$ and $x=2$ yields $g(0)=0$ and $g(2)=2$ respectively. We pass on to the key observations 
$$ \begin{array}{ll} g(x)+g(2-x)=2 & \text { for all } x \in \mathbb{R} \\ g(x+2)-g(x)=2 & \text { for all } x \in \mathbb{R} \end{array} $$ 
Replace $x$ by $1-x$ in (2), then change $x$ to $-x$ in the resulting equation. We obtain the relations $g(x)-g(-x)=C(g(1-x)-1), g(-x)-g(x)=C(g(1+x)-1)$. Then adding them up leads to $C(g(1-x)+g(1+x)-2)=0$. Thus $C \neq 0$ implies (3). Let $u, v$ be such that $u+v=1$. Apply (1) to the pairs $(u, v)$ and $(2-u, 2-v)$ : 
$$ g(1+u v)-g(1)=(g(u)-1)(g(v)-1), \quad g(3+u v)-g(3)=(g(2-u)-1)(g(2-v)-1) . $$ 
Observe that the last two equations have equal right-hand sides by (3). Hence $u+v=1$ implies 
$$ g(u v+3)-g(u v+1)=g(3)-g(1) . $$ 
Each $x \leq 5 / 4$ is expressible in the form $x=u v+1$ with $u+v=1$ (the quadratic function $t^{2}-t+(x-1)$ has real roots for $\left.x \leq 5 / 4\right)$. Hence $g(x+2)-g(x)=g(3)-g(1)$ whenever $x \leq 5 / 4$. Because $g(x)=x$ holds for $x=0,1,2$, setting $x=0$ yields $g(3)=3$. This proves (4) for $x \leq 5 / 4$. If $x>5 / 4$ then $-x<5 / 4$ and so $g(2-x)-g(-x)=2$ by the above. On the other hand (3) gives $g(x)=2-g(2-x), g(x+2)=2-g(-x)$, so that $g(x+2)-g(x)=g(2-x)-g(-x)=2$. Thus (4) is true for all $x \in \mathbb{R}$. Now replace $x$ by $-x$ in (3) to obtain $g(-x)+g(2+x)=2$. In view of (4) this leads to $g(x)+g(-x)=0$, i. e. $g(-x)=-g(x)$ for all $x$. Taking this into account, we apply (1) to the pairs $(-x, y)$ and $(x,-y)$ : 
$g(1-x y)-g(-x+y)=(g(x)+1)(1-g(y)), \quad g(1-x y)-g(x-y)=(1-g(x))(g(y)+1)$. 
Adding up yields $g(1-x y)=1-g(x) g(y)$. Then $g(1+x y)=1+g(x) g(y)$ by (3). Now the original equation (1) takes the form $g(x+y)=g(x)+g(y)$. Hence $g$ is additive. By additivity $g(1+x y)=g(1)+g(x y)=1+g(x y)$; since $g(1+x y)=1+g(x) g(y)$ was shown above, we also have $g(x y)=g(x) g(y)$ ( $g$ is multiplicative). In particular $y=x$ gives $g\left(x^{2}\right)=g(x)^{2} \geq 0$ for all $x$, meaning that $g(x) \geq 0$ for $x \geq 0$. Since $g$ is additive and bounded from below on $[0,+\infty)$, it is linear; more exactly $g(x)=g(1) x=x$ for all $x \in \mathbb{R}$. In summary $f(x)=x-1, x \in \mathbb{R}$. It is straightforward that this function satisfies the requirements. Comment. There are functions that satisfy the given equation but vanish at -1 , for instance the constant function 0 and $f(x)=x^{2}-1, x \in \mathbb{R}$.