summation - How find this sum $f(x)=sum_{n=1}^{infty}frac{x^{n^2}}{n}$

First, Merry Christmas everyone!

Find this sum
$$f(x)=\sum_{n=1}^{\infty}\dfrac{x^{n^2}}{n},1>x\ge 0 \tag{1}$$

This problem is creat by Laurentiu Modan.and I can't see this solution.

I know this sum
$$\sum_{n=1}^{\infty}\dfrac{x^n}{n}=-\ln{(1-x)},-1\le x<1$$

and I know this
$$\sum_{n=1}^{\infty}x^{n^2}\approx \dfrac{\sqrt{\pi}}{2\sqrt{1-x}},x\to 1^{-}$$

But for $(1)$,I can't find it,Thank you.

This problem is from this