Given a real number $t\neq -1$. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that 
\[(t+1)f(1+xy)-f(x+y)=f(x+1)f(y+1)\]
for all $x,y\in\mathbb{R}$.
1. Let \( P(x, y) \) be the assertion of the given functional equation:
   \[
   (t+1)f(1+xy) - f(x+y) = f(x+1)f(y+1)
   \]
   for all \( x, y \in \mathbb{R} \).

2. Consider the constant solutions first. If \( f(x) = 0 \) for all \( x \), then the equation becomes:
   \[
   (t+1) \cdot 0 - 0 = 0 \cdot 0 \implies 0 = 0
   \]
   which is true. Thus, \( f(x) = 0 \) is a solution. 

3. If \( f(x) = t \) for all \( x \), then the equation becomes:
   \[
   (t+1)t - t = t \cdot t \implies t^2 + t - t = t^2 \implies t^2 = t^2
   \]
   which is also true. Thus, \( f(x) = t \) is another solution.

4. Now, consider non-constant solutions. Assume \( f(1) = a \neq 0 \). Let \( a = f(1) \).

5. Substitute \( y = 0 \) into \( P(x, y) \):
   \[
   (t+1)f(1) - f(x) = f(x+1)f(1)
   \]
   Simplifying, we get:
   \[
   (t+1)a - f(x) = f(x+1)a
   \]
   Solving for \( f(x+1) \):
   \[
   f(x+1) = \frac{(t+1)a - f(x)}{a}
   \]

6. Substitute \( y = 1 \) into \( P(x, y) \):
   \[
   (t+1)f(1+x) - f(x+1) = f(x+1)f(2)
   \]
   Using \( f(x+1) = \frac{(t+1)a - f(x)}{a} \), we get:
   \[
   (t+1)f(1+x) - \frac{(t+1)a - f(x)}{a} = \frac{(t+1)a - f(x)}{a} f(2)
   \]

7. Let \( Q(x, y) \) be the new assertion:
   \[
   -a(t+1)f(xy) - f(x+y) = -a(t+1)f(x) - a(t+1)f(y) + a^2 f(x)f(y)
   \]

8. Substitute \( x = 0 \) into \( Q(x, y) \):
   \[
   -a(t+1)f(0) - f(y) = -a(t+1)f(0) - a(t+1)f(y) + a^2 f(0)f(y)
   \]
   Simplifying, we get:
   \[
   f(y) = a(t+1)f(y) - a^2 f(0)f(y)
   \]

9. Assume \( f(0) = b \). Then:
   \[
   f(y) = a(t+1)f(y) - a^2 b f(y)
   \]
   Simplifying, we get:
   \[
   f(y)(1 - a(t+1) + a^2 b) = 0
   \]
   Since \( f(y) \) is non-constant, we must have:
   \[
   1 - a(t+1) + a^2 b = 0
   \]

10. Substitute \( x = 1 \) into \( Q(x, y) \):
    \[
    -a(t+1)f(y+1) - f(1+y) = -a(t+1)f(1) - a(t+1)f(y) + a^2 f(1)f(y)
    \]

11. Let \( R(x, y) \) be the new assertion:
    \[
    f(x+y) + t + 2 = (f(x) + t + 2) + (f(y) + t + 2)
    \]
    for all \( x, y \neq 0 \).

12. Setting \( g(x) = \frac{f(x) + t + 2}{t+1} \), we get:
    \[
    g(x+y) = g(x) + g(y)
    \]
    and
    \[
    g(xy) = g(x)g(y)
    \]

13. The classical unique non-constant solution to these functional equations is \( g(x) = x \). Thus:
    \[
    \frac{f(x) + t + 2}{t+1} = x \implies f(x) = (t+1)x - (t+2)
    \]

14. Therefore, the non-constant solution is:
    \[
    f(x) = (t+1)x - (t+2)
    \]

The final answer is \( \boxed{f(x) = 0 \text{ or } f(x) = t \text{ or } f(x) = (t+1)x - (t+2)} \)