Take $y_1\ge y_2\ge \cdots\ge y_n$, so that \begin{align*} \sum_{i=1}^{n}\frac{y_i^2}{y_{i+1}^2-y_{i+1}y_{i+2}+y_{i+2}^2} &= \frac{y_{n-1}^2}{y_n^2-y_ny_1+y_1^2}+\sum_{i=0}^{n-2}\frac{y_i^2}{y_{i+1}^2+y_{i+2}(y_{i+2}-y_{i+1})} \\ &> \frac{y_{n-1}^2}{y_n^2-y_ny_1+y_1^2}+\frac{y_n^2}{y_1^2}+\frac{y_1^2}{y_2^2}+\cdots+\frac{y_{n-2}^2}{y_{n-1}^2} \\ &> \frac{y_{n-1}^2}{y_1^2}+\frac{y_1^2}{y_2^2}+\cdots+\frac{y_{n-2}^2}{y_{n-1}^2} \\ &\ge n-1, \end{align*}where $y_1\ge y_{n-1}\ge y_n\ge0$ implies that \[\frac{y_{n-1}^2}{y_n^2-y_ny_1+y_1^2}+\frac{y_n^2}{y_1^2}\ge\frac{y_{n-1}^2}{y_1^2} \Longleftrightarrow y_n^4+y_1y_n(y_{n-1}^2-y_n^2)+y_n^2(y_1^2-y_{n-1}^2)\ge0.\]Furthermore, when $y_1=\cdots=y_{n-1}=1$ and $y_n$ approaches $0$, the LHS approaches $n-1$ as well, so $M=n-1$. (We have $y_2-y_1=\cdots=y_{n-1}-y_{n-2}=0$ and $y_n(y_n-y_{n-1})=0$.)