If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the minimum value of $(x^2+y^2)^2$ If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the maximum value of $(x^2+y^2)^2$
My attempt is as follows:
$$(x-y)^2\ge 0$$
$$x^2+y^2\ge 2xy$$
$$2(x^2+y^2)\ge x^2+y^2+2xy$$
\begin{equation}
2(x^2+y^2)\ge (x+y)^2\tag{1}
\end{equation}
Solving the given equation:
$$x^2+2xy-y^2=6$$
$$(x+y)^2=2y^2+6$$
So putting the value of $(x+y)^2$ in equation $1$
$$2(x^2+y^2)\ge 2y^2+6$$
$$x^2+y^2\ge y^2+3$$
\begin{equation}
x^2\ge 3\tag{2}
\end{equation}
So $x\in \left(-\infty,-\sqrt{3}\right) \cup \left(\sqrt{3},\infty\right)$
But how to proceed from here?

The function to be minimized is the fourth power of the distance from the origin, so will have its extrema at the same points at which the distance is also an extremum. The constraint is a hyperbola, so the minimum distance occurs at its vertices, therefore this problem is equivalent to computing the fourth power of a semiaxis length of this hyperbola.  
One approach is to compute the eigenvalues of the matrix $\frac16\begin{bmatrix}1&1\\1&-1\end{bmatrix}$, which work out to be $\pm\frac1{3\sqrt2}$. We want the positive eigenvalue. It is the reciprocal of the square of the semi-transverse axis length, so the minimum value of $(x^2+y^2)^2$ on this hyperbola is $(3\sqrt2)^2=18$.
