Answer: $\dfrac{16}{3}$.
Let $A(a; 1/a)$, $B(b; 1/b)$, $C(c; 1/c)$, $D(d; 1/d)$ be the marked points (see Fig. 1). Since any vertical and any horizontal line meets the hyperbola $y = 1/x$ at most at one point we see that the numbers $a$, $b$, $c$, $d$ are pairwise distinct. Since the opposite sides of the parallelogram are equal, we have
$$\begin{aligned} AB = CD &\iff (b-a)^2 + (1/b - 1/a)^2 = (d-c)^2 + (1/d - 1/c)^2 \ &\iff (b-a)^2 \left(1 + \frac{1}{(ab)^2}\right) = (d-c)^2 \left(1 + \frac{1}{(cd)^2}\right). \quad (1) \end{aligned}$$
Moreover, since opposite sides of the parallelogram are equal and parallel we see that their projections on any line are equal, in particular, their projections on $Ox$-axis are equal, i.e. $|b-a| = |d-c|$. Then from (1) it follows that
$$1 + \frac{1}{a^2 b^2} = 1 + \frac{1}{c^2 d^2} \implies a^2 b^2 = c^2 d^2.$$
Similarly, from $BC = AD$ we obtain $b^2c^2 = a^2d^2$. Therefore, $a^2 = c^2$ and $b^2 = d^2$, and so, taking into consideration $a \neq c$ and $b \neq d$, we obtain $a = -c$, $d = -b$. Without loss of generality we suppose that $a < 0 < b$, and then $b < c$. Since $AB = 2BC$ we have
$$(b+c)^2 \left(1+\frac{1}{(bc)^2}\right) = (b-a)^2 \left(1+\frac{1}{(ab)^2}\right) = AB^2 = (2BC)^2 =$$
$$= 4(c-b)^2 \left(1 + \frac{1}{(bc)^2}\right),$$
so $(b+c)^2 = 4(b-c)^2$. Then $b+c = 2(c-b)$, so we have $c = 3b$.

Fig. 1

Fig. 2
Consider the pentagon $B_1BCC_2O$ (see Fig. 2). Since $B_1(0, 1/b)$, $B_2(b, 0)$, $C_1(0, 1/c)$, $C_2(c, 0)$ we have
$$\begin{align*} S(B_1BCC_2O) &= S(B_1BB_2O) + S(BCC_2B_2) = b \cdot \frac{1}{b} + \frac{1/b + 1/c}{2} \cdot (c-b) = \\ &= [c=3b] = 1 + \frac{4}{3b} \cdot \frac{1}{2} \cdot 2b = \frac{7}{3}. \end{align*}$$
On the other hand,
$$\begin{align*} S(B_1BCC_2O) &= S(OB_1B) + S(OBC) + S(OCC_2) = \\ &= \frac{1}{2} \cdot \frac{1}{b} \cdot b + S(OBC) + \frac{1}{2} \cdot \frac{1}{c} \cdot c = 1 + S(OBC), \end{align*}$$
so $S(OBC) = \frac{4}{3}$. Since $S(ABCD) = 4S(OBC)$ the required area of the parallelogram $ABCD$ is equal $\dfrac{16}{3}$.