Find all monotonic functions $u:\mathbb{R}\rightarrow\mathbb{R}$ which have the property that there exists a strictly monotonic function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[f(x+y)=f(x)u(x)+f(y) \]
for all $x,y\in\mathbb{R}$.

[i]Vasile Pop[/i]
1. **Restate the problem with the corrected conditions:**
   We need to find all functions \( u: \mathbb{R} \to \mathbb{R} \) such that there exists a strictly monotonic function \( f: \mathbb{R} \to \mathbb{R} \) satisfying the functional equation:
   \[
   f(x+y) = f(x)u(y) + f(y)
   \]
   for all \( x, y \in \mathbb{R} \).

2. **Define the assertion \( P(x, y) \):**
   Let \( P(x, y) \) denote the assertion of the given functional equation:
   \[
   P(x, y): f(x+y) = f(x)u(y) + f(y)
   \]

3. **Analyze \( P(x-y, y) \):**
   Consider \( P(x-y, y) \):
   \[
   f(x) = f(x-y)u(y) + f(y)
   \]
   Rearrange to isolate \( f(x-y) \):
   \[
   f(x) - f(y) = f(x-y)u(y)
   \]

4. **Consider the case \( u \equiv 0 \):**
   If \( u \equiv 0 \), then:
   \[
   f(x+y) = f(y)
   \]
   This implies \( f \) is constant, which contradicts the strict monotonicity of \( f \). Therefore, \( u \) cannot be identically zero.

5. **Find \( f(0) \):**
   Suppose there exists \( x_0 \) such that \( u(x_0) \neq 0 \). Take \( x = y = x_0 \) in the equation:
   \[
   f(0) = f(0)u(x_0)
   \]
   Since \( u(x_0) \neq 0 \), we must have \( f(0) = 0 \).

6. **Determine \( u(0) \):**
   Take \( y = 0 \) in the equation:
   \[
   f(x) = f(x)u(0)
   \]
   Since \( f(x) \neq 0 \) for \( x \neq 0 \) (due to strict monotonicity), we must have \( u(0) = 1 \).

7. **Analyze \( P(y, x) \):**
   Consider \( P(y, x) \):
   \[
   f(x+y) = f(x)u(y) + f(y) = f(y)u(x) + f(x)
   \]
   This implies:
   \[
   f(x)(u(y) - 1) = f(y)(u(x) - 1)
   \]
   Since \( f(x) \neq 0 \) for \( x \neq 0 \), we can write:
   \[
   \frac{u(x) - 1}{f(x)} = \frac{u(y) - 1}{f(y)}
   \]
   Let \( C \) be the constant of proportionality:
   \[
   u(x) - 1 = Cf(x)
   \]
   for all \( x \neq 0 \). For \( x = 0 \), we have \( u(0) - 1 = 0 = Cf(0) \), which is consistent.

8. **Substitute \( u(x) = 1 + Cf(x) \) into the original equation:**
   \[
   f(x+y) = f(x)(1 + Cf(y)) + f(y)
   \]
   Simplify:
   \[
   f(x+y) = f(x) + Cf(x)f(y) + f(y)
   \]

9. **Determine the form of \( u(x) \):**
   From \( u(x+y) = u(x)u(y) \), we recognize this as the functional equation for an exponential function. Since \( u \) is monotonic, we can take logarithms:
   \[
   \log u(x+y) = \log u(x) + \log u(y)
   \]
   Let \( g(x) = \log u(x) \). Then \( g(x+y) = g(x) + g(y) \), which is Cauchy's functional equation. The only monotonic solutions are of the form \( g(x) = kx \), so:
   \[
   u(x) = e^{kx}
   \]

10. **Verify the solution:**
    Substitute \( u(x) = e^{kx} \) back into the original equation and verify that it satisfies the conditions.

The final answer is \( \boxed{ u(x) = e^{kx} } \) for some constant \( k \).