1. Interpret the curve

The given curve is $x^2y^2=1 \,\Longrightarrow\, (xy)^2=1 \Longrightarrow xy=\pm 1.$ So the vertices and the midpoints of the square all lie on the pair of straight lines $xy=1 \quad \text{and} \quad xy=-1.$ These are the two rectangular hyperbola branches, equivalently the four lines after rotating axes.

2. Use symmetry of the square

Let the center of the square be at the origin, and let its vertices be represented as $A=\vec u+\vec v,\quad B=-\vec u+\vec v,\quad C=-\vec u-\vec v,\quad D=\vec u-\vec v,$ where $\vec u$ and $\vec v$ are perpendicular vectors of equal length.

Then the midpoints of the sides are $M_1=\vec v,\quad M_2=-\vec u,\quad M_3=-\vec v,\quad M_4=\vec u.$ So the four midpoints are simply at $\pm \vec u, \pm \vec v$.

Let $\vec u=(p,q), \qquad \vec v=(r,s).$ Since $\vec u \perp \vec v$ and $|\vec u|=|\vec v|$, we have $pr+qs=0, \qquad p^2+q^2=r^2+s^2.$

3. Condition from the midpoints lying on $x^2y^2=1$

Each midpoint lies on the curve, so for $\vec u=(p,q)$ and $\vec v=(r,s)$, $p^2q^2=1 \implies pq=\pm 1,$ $r^2s^2=1 \implies rs=\pm 1.$ Thus, $pq=\pm 1, \qquad rs=\pm 1.$

4. Parametrize using perpendicular equal vectors

Because $\vec u$ and $\vec v$ are perpendicular and equal in length, one can write $\vec v = (-q,p) \quad \text{or} \quad (q,-p).$ Take $\vec v=(-q,p).$ Then indeed $\vec u\cdot \vec v = p(-q)+q(p)=0,$ and $|\vec v|^2=q^2+p^2=|\vec u|^2.$

Now midpoint condition on $\vec v$ gives $rs=(-q)(p)=-pq=\pm 1,$ which is automatically satisfied if $pq=\mp 1$. So it is enough to take $pq=\pm 1.$

5. Write the vertices and impose the vertex condition

The vertices are $A=\vec u+\vec v=(p-q,\, q+p),$ $B=-\vec u+\vec v=(-p-q,\, -q+p),$ $C=-A,\quad D=-B.$

Since vertices also lie on $x^2y^2=1$, we need for vertex $A$: $[(p-q)(p+q)]^2=1.$ But $(p-q)(p+q)=p^2-q^2.$ Hence $(p^2-q^2)^2=1 \implies p^2-q^2=\pm 1.$

Also, from midpoint condition, $pq=\pm 1.$

So we must solve $pq=\pm 1, \qquad p^2-q^2=\pm 1.$

6. Solve for $p^2+q^2$

Let $a=p^2, \quad b=q^2.$ Then $ab=1,$ and $a-b=\pm 1.$ Squaring, $(a-b)^2=1 \implies a^2+b^2-2ab=1.$ Since $ab=1$, $a^2+b^2-2=1 \implies a^2+b^2=3.$ Now $(a+b)^2=a^2+b^2+2ab=3+2=5,$ so $a+b=\sqrt{5}.$ Thus $p^2+q^2=\sqrt{5}.$

7. Find side length and area of the square

The vertices are $\pm \vec u \pm \vec v$. Hence a side, for example $AB = |(\vec u+\vec v)-(-\vec u+\vec v)| = |2\vec u| = 2|\vec u|.$ Therefore side length is $s=2\sqrt{p^2+q^2}=2\sqrt{\sqrt{5}}.$ So area is $\text{Area}=s^2=4(p^2+q^2)=4\sqrt{5}.$ Hence the square of the area is $(\text{Area})^2=(4\sqrt{5})^2=80.$

8. Final answer

Therefore, the square of the area of $ABCD$ is $\boxed{80}.$