Find the value of K. Use of l Hospital's rule and expansion is not allowed.Let...

(Preamble: Apart from addition and multiplication the only genuine two-ary function allowed in print is the general power function:
$\begin{array}{}\text{(1)}& (x,y)\mapsto x^y:=\exp (y\log x)(x>0,y\in \mathbb{C}).\end{array}$
The function $(x,y)\mapsto {\log }_{x}y$ where both $x$ and $y$ are variables is not of this kind. When such a thing should really come up in practice one would have to rewrite it in such a way that only functions of one variable and the "allowed" operations appear. It is true that we sometimes write ${\log }_{a}y$. But the $a>1$ here is not a variable; it is a constant scaling parameter chosen in a most suitable way for the problem at hand.)
So after a lot of mind bending I'm interpreting your problem as follows: For $0<|x|\ll 1$ one has $0<\cos (3x)<1$. On the other hand
$\cos (2ix)=\frac{e^{-2x}+e^{2x}}{2}=\cosh (2x)(x\in \mathbb{R}).$
Therefore the equation
${(\cos (3x))}^y=\cos (2ix)=\cosh (2x)$
has a real solution $y$ for such $x$; and according to $(1)$ this $y$ is given by
$\begin{array}{}\text{(2)}& y=\frac{\log (\cosh (2x))}{\log (\cos (3x))}=:f(x)(0<|x|\ll 1).\end{array}$
The right hand side of $(2)$ is the correct expression of the function $f$ your teacher had in mind.
Now it comes to computing $\underset{x\to 0}{lim}f(x)$. Since both de l'Hôpital's rule and expansion are forbidden I have to stop here. It is unclear to me how this limit can be computed without using the analytical properties of the involved functions somehow.