Finding the Function

Differentiating the given relation with respect to $x$ we get

$\displaystyle 3f'\left(x+y\right)=f'\left(x\right)f\left(y\right)$

Putting $x=0$ we get

$\displaystyle f'\left(y\right)=2f\left(y\right)$

This Differential Equation can be solved easily and using $f\left(0\right)=3$ we finally get the function

$\displaystyle f\left(x\right)=3e^{2x}$

So, $f\left(1\right)=3e^2=22.167168...$

Given that: $f$ is dufferentiable over $x, y \in \mathbb R$ , $f(0)= 3$ , $f'(0)=6$ , and that

$\begin{aligned} f(x+y) & = \frac 13 f(x) f(y) & \small \color{#3D99F6} \text{Putting }y=x \\ f(2x) & = \frac 13 (f(x))^2 & \small \color{#3D99F6} \text{Differentiate both sides} \\ 2f'(2x) & = \frac 23 f(x) f'(x) \\ f'(2x) & = \frac 13 f(x) f'(x) & \small \color{#3D99F6} \text{Note that }f(2x) = \frac 13 (f(x))^2 \\ f'(2x) & = \frac {f(2x) f'(x)}{f(x)} & \small \color{#3D99F6} \text{Divide both sides by }f(2x) \\ \frac {f'(2x)}{f(2x)} & = \frac {f'(x)}{f(x)} = \color{#3D99F6} k & \small \color{#3D99F6} \text{where }k \text{ is a constant.} \\ \implies f'(x) & = k f(x) & \small \color{#3D99F6} \text{Solving the differential equation} \\ f(x) & = {\color{#3D99F6}a} e^{kx} & \small \color{#3D99F6} \text{where }a \text{ is a constant.} \\ f(0) & = a = 3 \\ f'(x) & = 3ke^{kx} & \small \color{#3D99F6} \text{As }f'(0) = 6 \implies k = 2 \\ \implies f(x) & = 3e^{2x} \\ f(1) & = 3e^2 \approx \boxed{22.167} \end{aligned}$