finding eigenvalues and eigenvectors 
Find the eigenvalues and eigenvectors 
  \begin{pmatrix}
-7 & 0 & -8 \\
2 & 1 & 2 \\
6 & 0 & 7
\end{pmatrix}

$\begin{bmatrix}
-7-x & 0 & -8 \\
2 & 1-x & 2 \\
6 & 0 & 7-x
\end{bmatrix}=(1-x) \cdot \begin{bmatrix}
-7-x &  & -8 \\
6 &  & 7-x
\end{bmatrix}=(1-x)*[-(7+x)(7-x)+48]=(1-x)(-49+x^2+48)=(1-x)(-1+x^2)=-1+x^2+x-x^3$
$-1+x^2+x-x^3=0$
$x_{1}=1$
$x_{2}=-1$
for $x_{1}=1$ we have $\begin{pmatrix}
-8 & 0 & -8 \\
2 & 0 & 2 \\
6 & 0 & 6
\end{pmatrix}\Rightarrow\begin{pmatrix}
1 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}$ so we have $v_1=\begin{pmatrix}
1   \\
0   \\
1  
\end{pmatrix}$, $v_2=\begin{pmatrix}
0   \\
1   \\
0
\end{pmatrix}$
for $x_{1}=-1$ we have $\begin{pmatrix}
-6 & 0 & -8 \\
2 & 2 & 2 \\
6 & 0 & 8
\end{pmatrix}\Rightarrow\begin{pmatrix}
1 & 1 & 1 \\
0 & 6 & -2 \\
0 & 0 & 0
\end{pmatrix}$ so we have $x+y+z=0$ and $6y-2z=0$, $6y=2z\rightarrow y=\frac{1}{3}z$ and $x+\frac{1}{3}z+z=0\rightarrow x=-\frac{4}{3}z $
so we have $v_1\begin{pmatrix}
-\frac{4}{3}   \\
\frac{1}{3}   \\
1
\end{pmatrix}$
Where did I get it wrong?

As per @pyrazolam's comment, your answer is correct.
For a given eigenvalue, there are an infinite number of associated eigenvectors; they're all (non-zero) multiples of one another. "$\mathbf{v}$ is an eigenvector for eigenvalue $\lambda$ of matrix $A$" means $A\mathbf{v}=\lambda\mathbf{v}$.  If we consider the vector $k\mathbf{v}$ instead, it's still an eigenvector for $\lambda$: $A(k\mathbf{v}) = k(A\mathbf{v}) = k(\lambda\mathbf{v})=\lambda(k\mathbf{v})$.
You got $\begin{bmatrix}-\frac43\\\frac13\\1\end{bmatrix}$ as an eigenvector for eigenvalue $-1$. You could multiply it by $3$ to get $\begin{bmatrix}-4\\1\\3\end{bmatrix}$  (still an eigenvector for $\lambda=-1$), and if you divide by $-4$ you get $\begin{bmatrix}1\\-\frac14\\-\frac34\end{bmatrix}$ which is the equally value eigenvector given by your link to the matrix calculator page.
