Just don't get bored

$f^2(x)=\dfrac{f(x)}{f^1(x)}.......(1)$

$\therefore f^3(x) = \dfrac{f^1(x)}{f^1(x)} - \dfrac{f(x)*f^2(x)}{(f^1(x))^2} =1 - \dfrac{(f(x))^2}{(f^1(x))^3}.......\space from(1)$

$\therefore f^4(x) = \dfrac{-2f(x)}{(f^1(x))^2}+\dfrac{3(f(x))^3}{(f^1(x))^5}$

$\therefore f^5(x) = \dfrac{-2}{f^1(x)}+\dfrac{4*(f(x))^2}{(f^1(x))^4}+\dfrac{9*(f(x))^2}{(f^1(x))^4}-\dfrac{15(f(x))^4}{(f^1(x))^7}$

In this way if you will go then you will get,

$f^7(x) = \dfrac{2}{(f^1(x))^2}+\dfrac{18}{(f^1(x))^2}+\dfrac{8}{(f^1(x))^2}+k*f(x)$

So, $f(a)=0$

$f^7(a) = \dfrac{2}{(f^1(a))^2}+\dfrac{18}{(f^1(a))^2}+\dfrac{8}{(f^1(a))^2}+k*f(a)$

$f^7(a) = \dfrac{2}{(f^1(a))^2}+\dfrac{18}{(f^1(a))^2}+\dfrac{8}{(f^1(a))^2}+0$

$f^7(a)*(f^1(a))^2 =\boxed{28}$

Here the noticeable thing is that, $f^k(a) = 0$ if k is an even integer.