The given matrix has three linearly independent eigenvectors, then $x+y=0$. The question asked is  ::
If the matrix
$$
A=\left(\begin{array}{lll}
0 & 0 & 1 \\
x & 1 & y \\
1 & 0 & 0
\end{array}\right)
$$
has three linearly independent eigenvectors, then show that $x+y=0$.
solving for eigenvalues from the characteristic polynomial:
$$\left|\begin{matrix}
0-\lambda & 0 & 1 \\
x & 1-\lambda & y \\
1 & 0 & 0-\lambda
\end{matrix}\right| =-λ^3+λ^2+λ-1$$
$$=-(λ-1)*(λ^2-1)=-(λ-1)*(λ-1)=-(λ-1)^2*(λ+1)$$
So eigenvalues are $λ_1=1$ and $λ_2=-1$, Independent of the values of $x$ and $y$.
Now solving for eigenvectors I got
$\left(\begin{matrix}
0 \\
1 \\
0
\end{matrix}\right)$ and $\left(\begin{matrix}
-1 \\
\frac{x-y}{2} \\
1
\end{matrix}\right)$
From here how to show that if there are three linearly independent eigenvectors, then show that $x+y=0$.

Your eigenvectors are correct, however, there is a possibility that there are $1$ or $2$ eigenvectors corresponding to the eigenvalue $\lambda = 1$, depending on the values of $x$ and $y$.
Let us find the missing one. We need to solve the homogeneous system $(A-I)X = 0$:
$$\begin{pmatrix}-1&0&1\\x&0&y\\1&0&-1\end{pmatrix}\sim \begin{pmatrix}1&0&-1\\0&0&0\\0&0&x+y\end{pmatrix}.$$
There are $2$ cases, $x+y \neq 0$ or $x+y= 0$.
If $x+y\neq 0$, the system is of rank $2$, and therefore $(0, 1, 0)$ is the only solution (up to scalar).
If $x+y = 0$, the system is of rank $1$, and therefore we have $2$ linearly independent solutions: $(0,1,0)$ and $(1,0,1)$.
Note that the converses are true as well.
If we count the third eigenvector, corresponding to $\lambda = -1$, that you found, we can sum up all of this in the following way:

*

*$A$ has exactly $2$ linearly independent eigenvectors if and only if $A-I$ is of rank $2$, if and only if $x+y\neq 0$.

*$A$ has exactly $3$ linearly independent eigenvectors if and only if $A-I$ is of rank $1$, if and only if $x+y= 0$.

