$$
\because \quad f(x+y)=f(x) f(y), \forall x, y \in R
$$
Put $x=y=1$, we get
$$
\begin{aligned}
& f(2)=f(1) \cdot f(1)=9 \quad[\because f(2)=9] \\
& \Rightarrow \quad f(1)^2=9 \Rightarrow f(1)=3 \\
&
\end{aligned}
$$
Now, put $x=2$ and $y=1$ in Eq. (i), we get
$$
f(3)=f(2) \cdot f(1)=3^2 \cdot 3=3^3
$$
Now, put $x=3$ and $y=1$ in Eq. (i), we get
$$
f(4)=f(3) \cdot f(1)=3^3 \cdot 3=3^4
$$
Again, put $x=4$ and $y=2$ in Eq. (i), we get
$$
f(6)=f(4) \cdot f(2)=3^4 \cdot 3^2=3^6
$$
Alternative Method
We have,
$$
\begin{aligned}
f(x+y) & =f(x) f(y), \forall x, y \in R \\
f(2) & =9
\end{aligned}
$$
and
Now,
$$
\begin{aligned}
f(1+1) & =f(1) \cdot f(1) \\
f(2) & =\{f(1)\}^2 \\
\{f(1)\} & =\sqrt{\{f(2)\}}
\end{aligned}
$$
$$
\Rightarrow
$$
$$
\Rightarrow
$$
Now,
$$
\{f(1)\}=\sqrt{\{f(2)\}}
$$
$$
f(6)=f(1+1+1+1+1+1)
$$
$$
\begin{aligned}
& =f(1) \cdot f(1) \cdot f(1) \cdot f(1) \cdot f(1) \cdot f(1) \\
& =\{f(1)\}^6 \quad \text { [using Eq. (ii) }
\end{aligned}
$$
[using Eq. (ii)]
$$
=[\sqrt{\{f(2)\}}]^6=[f(2)]^3
$$
$=(9)^3$
[using Eq. (i)]
$=(3)^6$