What is the the function $f(x)$ has the property $\sum_{n=1}^x f(x^n)=x^2$

Question

What is the the function $f(z)$ has the property $\sum_{n=0}^z f(z^n)=z^2$ when $z$ is a natural number bigger than $0$, and the function $f(z)$ is differentiable everywhere. $$f(1)+f(1)=1$$ $$f(1)+f(2)+f(4)=4$$ $$f(1)+f(3)+f(9)+f(27)=9$$ $$\dots$$ $$f(1)+f(z)+f(z^2)+f(z^3)+\dots+f(z^z)=z^2$$ I've found that $f(1)=\frac{1}{2}$ but the others I can't find a value to.