Answer: No.

Solution 1: Let $f(x) = e^{\sin \ln x}$. Then
$$(f(cx))^2 = (e^{\sin \ln(cx)})^2 = e^{2\sin(\ln cx + \ln c)}, \\ f(x)f(c^2x) = e^{\sin \ln x} \cdot e^{\sin \ln(c^2x)} = e^{\sin \ln x + \sin(\ln x + 2\ln c)}.$$
Taking $c = e^{2\pi}$, we get
$$\begin{aligned} 2 \sin (\ln x + \ln c) &= 2 \sin (\ln x + 2\pi) = 2 \sin \ln x = \sin \ln x + \sin \ln x \\ &= \sin \ln x + \sin (\ln x + 4\pi) \\ &= \sin \ln x + \sin (\ln x + 2\ln c), \end{aligned}$$
so the desired condition is satisfied. However, taking $c = e^{\pi/2}$ and $x = 1$, we obtain $2\sin(\ln 1 + \ln e^{\pi/2}) = 2\sin(\pi/2) = 2$, whereas $\sin \ln 1 + \sin(\ln 1 + 2\ln e^{\pi/2}) = \sin 0 + \sin \pi = 0$. Hence the equality does not hold for $c = e^{\pi/2}$ and $x = 1$.

Solution 2: Let
$$f(x) = \begin{cases} 2, & \text{if } x \text{ is rational}, \\ 1, & \text{if } x \text{ is irrational}. \end{cases}$$
For all positive rational $c$ we have $f(cx) = f(x)$, as $x$ and $cx$ are either both rational or both irrational. So $(f(cx))^2 = (f(x))^2$ and $f(x)f(c^2x) = f(x)f(cx) = (f(x))^2$. Hence the desired equality holds for all positive rational $c$. However, taking $c = \sqrt{2}$ and $x = 1$, we obtain $f(cx) = f(\sqrt{2}) = 1$, whereas $f(x)f(c^2x) = f(1)f(2) = 2 \cdot 2 = 4$. So the equality does not hold.