Find all functions $u:R\rightarrow{R}$ for which there exists a strictly monotonic function $f:R\rightarrow{R}$ such that $f(x+y)=f(x)u(y)+f(y)$
for all $x,y\in{\mathbb{R}}$
1. Let \( P(x, y) \) be the assertion \( f(x+y) = f(x)u(y) + f(y) \).
2. Consider \( P(0, x) \):
   \[
   f(0 + x) = f(0)u(x) + f(x) \implies f(x) = f(0)u(x) + f(x)
   \]
   Subtract \( f(x) \) from both sides:
   \[
   0 = f(0)u(x) \implies f(0)u(x) = 0
   \]
   Since this must hold for all \( x \), we have \( f(0) = 0 \) (otherwise, \( u(x) = 0 \) for all \( x \), which contradicts the strict monotonicity of \( f \)).

3. Consider \( P(x, 0) \):
   \[
   f(x + 0) = f(x)u(0) + f(0) \implies f(x) = f(x)u(0)
   \]
   Since \( f(x) \) is not identically zero, we must have:
   \[
   u(0) = 1
   \]

4. If \( u(t) = 0 \) for some \( t \), then \( P(x, t) \) implies:
   \[
   f(x + t) = f(x)u(t) + f(t) \implies f(x + t) = f(t)
   \]
   This would mean \( f \) is constant, contradicting the strict monotonicity of \( f \). Therefore, \( u(x) \neq 0 \) for all \( x \).

5. Subtracting \( P(x, y) \) from \( P(y, x) \), we get:
   \[
   f(x)u(y) + f(y) = f(y)u(x) + f(x)
   \]
   Rearranging, we obtain:
   \[
   f(x)(u(y) - 1) = f(y)(u(x) - 1)
   \]
   Let this be assertion \( Q(x, y) \).

6. If \( u(t) = 1 \) for some \( t \neq 0 \), then \( Q(x, t) \) implies:
   \[
   f(x)(1 - 1) = f(t)(u(x) - 1) \implies 0 = f(t)(u(x) - 1)
   \]
   Since \( f(t) \neq 0 \), we must have \( u(x) = 1 \) for all \( x \). This is indeed a solution, as choosing \( f(x) = x \) satisfies the original equation.

7. If \( u(x) = 1 \iff x = 0 \), then \( Q(x, 1) \) implies:
   \[
   f(x) = c(u(x) - 1)
   \]
   for some constant \( c = \frac{f(1)}{u(1) - 1} \neq 0 \).

8. Plugging this into the original equation, we get:
   \[
   f(x + y) = c(u(x + y) - 1)
   \]
   and
   \[
   f(x)u(y) + f(y) = c(u(x) - 1)u(y) + c(u(y) - 1)
   \]
   Equating these, we get:
   \[
   c(u(x + y) - 1) = c(u(x)u(y) - u(y) + u(y) - 1)
   \]
   Simplifying, we get:
   \[
   u(x + y) = u(x)u(y)
   \]
   Since \( u(x) \neq 0 \) for all \( x \), this implies \( u(x) > 0 \) for all \( x \).

9. Therefore, \( \ln u(x) \) is additive:
   \[
   \ln u(x + y) = \ln u(x) + \ln u(y)
   \]
   Hence, \( u(x) = e^{g(x)} \) where \( g(x) \) is an additive function.

10. If \( g(x) \) is not linear, the graph of \( g(x) \) is dense in \( \mathbb{R}^2 \), and so the graph of \( f(x) \) is dense in \( \mathbb{R} \times (-c, +\infty) \) (or \( (-\infty, -c) \) if \( c < 0 \)), which contradicts the strict monotonicity of \( f \).

11. Therefore, \( g(x) \) must be linear, and we have:
   \[
   u(x) = e^{ax} \quad \forall x
   \]
   which indeed is a solution, whatever \( a \in \mathbb{R} \), choosing for example \( f(x) = e^{ax} - 1 \) if \( a \neq 0 \) or \( f(x) = x \) if \( a = 0 \).

The final answer is \( \boxed{u(x) = e^{ax} \forall x} \).