Explanation

(a)
Since, $f^{'}(0) = \lim_{x \rightarrow 0}\frac{f(x) - f(0)}{x - 0} = 10$
$\Rightarrow \ \ \lim_{h \rightarrow 0}\ \ \frac{f(0 + h) - f(0)}{h} = 10$
$\Rightarrow f(0)\left( \lim_{h \rightarrow 0}\frac{f(h) - 1}{h} \right) = 10$ ...(i)
$\lbrack\because\ \ \ f(0 + h) = f(0)f(h),\ given\rbrack\$
Now, $f(0) = f(0)f(0)$
$\Rightarrow f(0) = 1$
$\therefore$ From Eq. (i)
$\lim_{h \rightarrow 0}\ \frac{f(h) - 1}{h} = 10$ ...(ii)
Now, $f^{'}(6) = \lim_{h \rightarrow 0}\frac{f(6 + h) - f(6)}{h}$
$= \lim_{x \rightarrow 0}\ \left( \frac{f(h) - 1}{h} \right)f(6)$ [from Eq. (ii)]
$= 10 \times 3 = 30$