How to prove $\sum_{k=1}^\infty\frac{k^k}{k!}x^k=\frac{1}{2}$ where $x=\frac{1}{3}e^{-1/3}$

Idea, too long for a comment: use the Lambert function. $$W(x) = \sum_{n=1}^\infty\frac{(-n)^{n-1}}{n!}x^n,$$ $$W'(x) = \sum_{n=1}^\infty\frac{(-1)^{n-1}n^n}{n!}x^{n-1} = \sum_{n=1}^\infty\frac{n^n}{n!}(-x)^{n-1}$$ $$\cdots$$