I know the answer to the functional equation
$$\frac{f(x)}{C}=f(Cx)$$
is $f(x)=\frac{C_{2}}{x}$ , where $C$ and $C_{2}$ are constants.
How can we solve this algebraically using differential equation theoretical techniques (without "guessing" intuitionally)?
On
Since $f(Cx)=\frac{f(x)}{C}$, assuming $C\not\in \{0,1\}$, otherwise the functional equation is trivial or undefined, then choosing $x=C^k$ yields \begin{align} f(C^{k+1}) = f(CC^{k}) = \frac{f(C^k)}{C} = \frac{f(C^{k-1})}{C^2} = \dots = \frac{f(1)}{C^{k+1}}. \end{align} This gives all the values of $f$ on $\{ C^k : k\in\mathbb{Z}\}$ by induction (in positive and negative directions). If we choose arbitrary values for $f$ on the set $[1,C)$, then we can use the above method to calculate all the values of $f$ on $\{ aC^k : k\in\mathbb{Z},\,a\in[1,C)\}=\mathbb{R}_+$. There are therefore infinitely many solutions. We can choose two such functions $f_1,f_2$ and construct a function on $\mathbb{R}_{\neq 0}$ which satifies the functional equation: $f(x) = f_1(x)$ for $x>0$, and $f(x)=f_2(-x)$ for $x<0$ satifies the functional equation.
If we assume that we only find functions $f$ which are never equal zero and continuously differentiable on their domain, then, we can think as follows:
Put $g(x)=\frac{1}{f(x)}$. Then, our functional equation can be rewritten $$Cg(x)=g(Cx)$$
Differentiate this equation and obtain
$$g^{\prime}(x)=g^{\prime}(Cx)$$
(Since C is not zero)
By induction we get
$$g^{\prime}(x)=g^{\prime}(C^{n}x)$$ for all $x$ (from the domain) and all natural numbers $n$.
Let $n\to \infty$ and using the continuity of $g^{\prime}(x)$ we deduce that $g^{\prime}(x)$ is constant. Hence $g(x)=ax+b$ is linear. From our original equation we find that constant $a$ may be arbitrary, but non-zero and $b=0$. So $f(x)=\frac{1}{ax}$.
$\mathbf {Remark:}$ We see that $0$ should not be in the domain of $f$.