Showing there's no closed-form: $\sum_{n=0}^\infty(-1)^n\frac{\cos^2({3^nx})}{3^n}$

Comment extended to a "not an answer" answer per request.

There is another possible form of typo $$\sum_{n=0}^\infty (-1)^n \frac{\cos^{\color{red}{3}}(3^n x)}{3^n}$$ which sums to a closed form.

Start from the triple angle formula for cosine, $$\cos(3\theta) = 4\cos^3\theta - 3\cos\theta \quad\iff\quad\cos^3\theta = \frac34\left[\cos\theta + \frac{\cos(3\theta)}{3}\right]$$ We have $$\begin{align} (-1)^n\frac{\cos^3(3^n x)}{3^n} &= (-1)^n \frac34\left[\frac{\cos(3^n x)}{3^n} + \frac{\cos(3^{n+1}(x)}{3^{n+1}}\right]\\ &= \frac34\left[ (-1)^n \frac{\cos(3^n x)}{3^n} - (-1)^{n+1} \frac{\cos(3^{n+1} x)}{3^{n+1}} \right]\end{align} $$ This allows us to turn the sum into a telescoping sum. The end result is $$\sum_{n=0}^\infty (-1)^n \frac{\cos^3(3^n x)}{3^n} = \frac34 \times (-1)^0 \frac{\cos(3^0 x)}{3^0} = \frac34 \cos(x)$$

Lets say $$f(x)=\sum_{n=0}^\infty(-1)^n\frac{\cos^2({3^nx})}{3^n}$$ Then $$f'(x)=-\sum_{n=0}^\infty(-1)^n\sin({2*3^nx})$$ Now $$\sin(t)=t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}+...$$ with $t=2*3^nx$ $$\sin(2*3^nx)=2*3^nx-\frac{(2*3^nx)^3}{3!}+\frac{(2*3^nx)^5}{5!}-\frac{(2*3^nx)^7}{7!}+...=2*3^nx-\frac{3^{3n}(2x)^3}{3!}+\frac{3^{5n}(2x)^5}{5!}-\frac{3^{7n}(2x)^7}{7!}+...$$ As $$\sum_{n=0}^\infty(-1)^n3^{mn}=\frac{1}{3^m+1} $$ The above relation becomes: $$f'(x)=-\sum_{k=0}^\infty\frac{(-1)^k(2x)^{2k+1}}{(1+3^{2k+1})(2k+1)!}$$ Not sure if this function has a closed form in terms of elementary functions.