singularity of analytic continuation of $f(z) = \sum_{n=1}^\infty \frac{z^n}{n^2}$

Note that when $|z|<1$, $$f'(z)=\sum_{n=1}^\infty\frac{z^{n-1}}{n}=-\frac{\log(1-z)}{z}.\tag{1}$$ Since the right hand side of $(1)$ has a unique singularity at $z=1$, it implies that on the unit circle, the analytic continuation of $f$ has a unique singularity at $z=1$.

For the general situation, note that for one complex variable, any (nonempty) open set in $\mathbb{C}$ is a domain of holomorphy, which implies that for any closed subset $S$ of the unit circle, there exists a holomorphic function $f$ defined on the unit disk, such that the collection of sigularities of the analytic continuation of $f$ on the unit circle is precisely $S$.

Remark: Just in case, $(1)$ follows from integrating $$(zf'(z))'=\sum_{n=0}^\infty z^n=\frac{1}{1-z}.$$

This is one of the polylogarithm, also $s=2$ case is called dilogarithm as GEdgar pointed out. The definition is $$ \operatorname{Li}_s(z) = \sum_{k=1}^\infty {z^k \over k^s} = z + {z^2 \over 2^s} + {z^3 \over 3^s} + \cdots \, $$ Your case it is $s=2$, so $\operatorname{Li}_2(z)$.

According to http://en.wikipedia.org/wiki/Polylogarithm, we have a integral representation of $\operatorname{Li}_s(z)$:

$$ \operatorname{Li}_{s}(z) = {1 \over \Gamma(s)} \int_0^\infty {t^{s-1} \over e^t/z-1} \,\mathrm{d}t \,. $$

So in your case $$ \operatorname{Li}_{2}(z) = \int_0^\infty {zt \over e^t - z} \,\mathrm{d}t \,.$$

This allows an analytic continuation to $z\in \mathbb{C}-[1,\infty)$.