Let
$ \textbf{A} = \pmatrix{0 & 1 & 0 \\ 3 & 0 & 2 \\ 0 & -1 & 0} $ then, expanding on the first row \begin{align} \det(\textbf{A} - \lambda \textbf{I}) = - \lambda \begin{vmatrix}-\lambda & 2 \\ -1 & -\lambda\end{vmatrix}-1 \begin{vmatrix}3 & 2 \\ 0 & -\lambda\end{vmatrix} &= -\lambda^3 - 2 \lambda +3\lambda \\ &= -\lambda(\lambda^2-1) \end{align} hence $ \lambda_1 = -1,\quad \lambda_2 = 0, \quad \lambda_3 = 1. $
To calculate the eigenvectors we need to solve $ (\textbf{A} - \lambda_i \textbf{I})\textbf{v}_i = 0 $
For $\lambda_1 = -1$ we have $ (\textbf{A} + \textbf{I})\textbf{v}_1 = \pmatrix{1 & 1 & 0 \\ 3 & 1 & 2 \\ 0 & -1 & 1} \pmatrix{v_{11} \\ v_{21} \\ v_{31} } = \pmatrix{0 \\ 0 \\ 0} $ and then $\textbf{v}_1 = (-1,1,1)^T$.
For $\lambda_2 = 0$ $ \textbf{A}\textbf{v}_2 = \pmatrix{0 & 1 & 0 \\ 3 & 0 & 2 \\ 0 & -1 & 0} \pmatrix{v_{12} \\ v_{22} \\ v_{32} } = \pmatrix{0 \\ 0 \\ 0} $ and $\textbf{v}_2 = \big(-\frac{2}{3}, 0 ,1\big)^T$
Finally, for $\lambda_3 = 1$ $ (\textbf{A} - \textbf{I})\textbf{v}_3 = \pmatrix{-1 & 1 & 0 \\ 3 & -1 & 2 \\ 0 & -1 & -1} \pmatrix{v_{13} \\ v_{23} \\ v_{33} } = \pmatrix{0 \\ 0 \\ 0} $ and $\textbf{v}_3 =(-1,-1,1)^T$ and the solution to the system is
$ \textbf{y} = c_1 \textbf{v}_1 e^{\lambda_1 t} + c_2 \textbf{v}_2 e^{\lambda_2 t} + c_3 \textbf{v}_3 e^{\lambda_3 t} $