$2x+y =\lambda2x, \\x+2y = \lambda 2y $
Once you have these equations, you should see that it is merely a problem of determining the eigenvalues $\lambda_1 , \lambda_2 $ of the matrix $M = \begin{pmatrix}
1 & 1/2 \\
1/2 & 1 \\
\end{pmatrix}$ . One of them will yield you the absolute minimum, and the other the absolute maximum.
This is done by solving the characteristic polynomial.
$
\left| \begin{array}{ccc}
1-\lambda & 1/2 \\
1/2 & 1-\lambda \\
\end{array} \right| = 0
$
Which yields:
$ (1-\lambda)^2 = 1/4 \rightarrow \lambda_1 =1/2 ,\lambda_2 =3/2$
Now you have either of the two cases:
$2x+ y = x \rightarrow (x,y)=(-1,1)*t\\
2x+ y = 3x \rightarrow (x,y) = (1,1)*t $
In order to have them with length of 8, you should find the relevant $t$ for which $|| (-t,t)||^2=8, ||(t,t)||^2=8$
Another way to think about your problem is that you have a quadratic form $ ax^2 + 2bxy + cy^2 $, where $ a = 1, b= 1/2 , c = 1$ . The maximal and minimal values of a quadratic form, under the constraint of constant length of $(x,y)$ is attained in its eigenvalues, and $(x,y)$ direction is the direction of the eigenvectors.