Assume that $f(z)=f(w)$, for $z\ne w$ and $\lvert z\rvert,\,\lvert w\rvert<\frac23$.
Then
$$
0=\sum_{n=1}^\infty\frac{z^n-w^n}{n^2}=(z-w)\sum_{n=1}^\infty\frac{z^{n-1+}z^{n-2}w+\cdots+w^{n-1}}{n^2}.
$$
Hence
$$
0=1+\sum_{n=2}^\infty\frac{z^{n-1+}z^{n-2}w+\cdots+w^{n-1}}{n^2}.
$$
But
$$
\sum_{n=2}^\infty\frac{\lvert z^{n-1+}z^{n-2}w+\cdots+w^{n-1}\rvert}{n^2}\le
\sum_{n=2}^\infty\frac{n(2/3)^{n-1}}{n^2}=\sum_{n=2}^\infty\frac{(2/3)^{n-1}}{n}=-\frac{3}{2}\big(1+\ln (1-2/3)\big)<1.
$$
Since, for $0<\lvert x\rvert<1$,
$$
\sum_{n=2}^\infty\frac{x^{n-1}}{n}=\frac{1}{x}\left(\sum_{n=1}^\infty\frac{x^{n}}{n}-1\right)=\frac{1}{x}(-\ln(1-x)-1).
$$
Note. This $f$ is univalent even in $D\big(\frac{5}{6}\big)$.