Estonian Math Competitions

Answer: No.

Solution 1: Let $f(x)=e^{\sin \ln x}$. Then $$\begin{gathered} (f(cx){)}^2=(e^{\sin \ln (cx)}{)}^2=e^{2\sin (\ln cx+\ln c)}, \\ f(x)f(c^{2}x)=e^{\sin \ln x}\cdot e^{\sin \ln (c^{2}x)}=e^{\sin \ln x+\sin (\ln x+2\ln c)}. \end{gathered}$$ Taking $c=e^{2\pi }$, we get $$\begin{array}{rl}2\sin (\ln x+\ln c)& =2\sin (\ln x+2\pi )=2\sin \ln x=\sin \ln x+\sin \ln x\\ & =\sin \ln x+\sin (\ln x+4\pi )\\ & =\sin \ln x+\sin (\ln x+2\ln c),\end{array}$$ 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 $$f(x)=\{\begin{array}{ll}2,& \text{if }x\text{ is rational},\\ 1,& \text{if }x\text{ is irrational}.\end{array}$$ 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 $f(x)f(c^{2}x)=f(x)f(cx)=(f(x){)}^2$. 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 $f(x)f(c^{2}x)=f(1)f(2)=2\cdot 2=4$. So the equality does not hold.