If a 3 by 3 matrix has three independent eigenvectors, then it can be "diagonalized". Here, the eigenvalues are, as you say, 0, 1, and 3. An eigenvector corresponding to eigenvalue 0 is <1, 1, 1>, an eigenvector corresponding to eigenvalue 1 is <-1, 0, 1>, and an eigenvector corresponding to eigenvalue 3 is <1, -2, 1>. The matrix having those eigenvectors as columns is
P= \begin{bmatrix}-1 & 1 & 1 \\ 0 & 1 & -2 \\ 1 & 1 & 1\end{bmatrix}
and its inverse is
P^{-1}= \begin{bmatrix}-\frac{1}{2} & 0 & \frac{1}{2}\\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{6} & -\frac{1}{3} & \frac{1}{6}\end{bmatrix}
and now it is easy to check that
D= P^{-1}AP= \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 3\end{bmatrix}
the diagonal matrix with the eigenvalues on the diagonal.
Now go back to Y'= AY, the differerential equation. We can multiply both sides, on the left, by P^{-1} to get P^{-1}Y'= P^{-1}AY. And, since P has only constant entries, and PP^{-1}= I, we can write that as (P^{-1}Y)'= P^{-1}APP^{-1}Y= D(P^{-1}Y) or, letting z= P^{-1}Y, z'= Dz which is just
\begin{bmatrix}z_1(t) \\ z_2(t) \\ z_3(t)\end{bmatrix}= \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 3\end{bmatrix}\begin{bmatrix}z_1(t) \\ z_2(t) \\ z_3(t)\end{bmatrix}
which is equivalent to the three separate equations z_1'= z_1, z_2'= 0, and z_3'= 3z_3 which, of course, have solutions z_1(t)= C_1e^t, z_2= C_2, z_3= C_3e^{3t}.
Now, because Z= P^{-1}Y, Y= PZ just multiply
Y= \begin{bmatrix}-1 & 1 & 1 \\ 0 & 1 & -2 \\ 1 & 1 & 1\end{bmatrix}\begin{bmatrix}C_1e^t \\ C_2 \\ C_3e^{3t}\end{bmatrix} to get
y_1(t)= -C_1e^t+ C_2+ C_3e^{3t}
y_2(t)= C_2- 2C_3e^{3t} and
y_3(t)= C_1e^t+ C_2+ C_3e^{3t}
