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?

Perform a parametrization of the form $$x = r \cos \theta, \quad y = r \sin \theta.$$  Then we seek to minimize $(x^2 + y^2)^2 = r^4$ subject to the constraint $$\begin{align*}
6 & = r^2 \cos^2 \theta + 2 r^2 \cos \theta \sin \theta - r^2 \sin^2 \theta \\
&= r^2 \left( \cos^2 \theta - \sin^2 \theta + 2\cos \theta \sin \theta\right) \\
&= r^2 \left( \cos 2\theta + \sin 2\theta \right).
\end{align*}$$  Therefore, $$r^2 = \frac{6}{\cos 2\theta + \sin 2\theta},$$ and on $\theta \in [0, 2\pi)$, the LHS is minimized when the denominator is maximized.  Rewriting this using an additional trigonometric identity, $$\cos 2\theta + \sin 2\theta = \sqrt{2} \left(\cos 2\theta \sin \frac{\pi}{4} + \sin 2\theta \cos \frac{\pi}{4}\right) = \sqrt{2} \sin \left( 2\theta + \frac{\pi}{4}\right).$$  So the maximum is attained whenever $2\theta + \frac{\pi}{4} = 2\pi k + \frac{\pi}{2}$ for some integer $k$; namely $$\theta \in \left\{\frac{\pi}{8}, \frac{9\pi}{8}\right\},$$ and the maximum value is $\sqrt{2}$; thus the minimum value is $$r^4 = \left( \frac{6}{\sqrt{2}} \right)^2 = 18.$$  It is a straightforward exercise to compute the set of $(x,y)$ values for which this minimum is attained.
