$$
\begin{aligned}
& \text { } f: R \rightarrow R \\
& f(x+y)=f(x) \cdot f(y), \forall x, y \in R \\
& f(x) \neq 0, \forall x \in R \\
& \text { and } f^{\prime}(0)=4 \text { and } f(6)=3 \\
& \because \quad f(x+y)=f(x) \cdot f(y) ...(i)\\
& \text { Put } x=0 \quad y=0 \\
& \quad f(0)=f(0) \cdot f(0) \\
& \Rightarrow f(0)[f(0)-1]=0 \\
& \quad f(0)=0 \text { and } f(0)-1=0
\end{aligned}
$$
not possible and $f(0)=1$
$$
\begin{aligned}
& f^{\prime}(0)=4 \\
& \Rightarrow \quad \lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=4 \\
& \Rightarrow \quad \lim _{h \rightarrow 0} \frac{(f(h)-1)}{h}=4 ...(ii)\\
&
\end{aligned}
$$
Now,
$$
\text { } \begin{aligned}
f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{f(x) \cdot f(h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} f(x) \cdot\left[\frac{f(h)-1}{h}\right] \\
& =f(x) \lim _{h \rightarrow 0} \frac{f(h)-1}{h} \\
& =f(x) \times 4[by Eq. (ii)] \\
\therefore \quad f^{\prime}(x) & =4 f(x) \\
\therefore \quad f^{\prime}(6) & =4 f(6)=4 \times 3=12
\end{aligned}
$$