Let $f(x) = -\sqrt{x_1 x_2}, x \in \mathbb{R^2_+}$.
First, I've proved that the function is convex and that it is positively homogeneous, so $sup_{x \in \mathbb{R^2_+}}\{\langle x, y\rangle - f(x)\} = +\infty$, for those $y \in \mathbb{R^2}$ for which exists $x \in \mathbb{R^2_+}$ such that $\langle x, y\rangle - f(x)\ \ge 0$. Thus $D_{f^c} = \{y \in \mathbb{R^2} | \langle x, y\rangle - f(x) \le 0\}$.
Then, I've noticed that if for $y \in \mathbb{R^2}$ is $\langle x, y\rangle - f(x)\ \le 0, \forall x \in \mathbb{R^2_+}$, then is also $sup_{x \in \mathbb{R^2_+}}\{\langle x, y\rangle - f(x)\} \le 0.$ But, for such $y \in \mathbb{R^2}$ is $\langle x, y\rangle - f(x) = 0$, when chosen $x = (0, 0).$ So, for such $y$ is $f^c(y) = 0$.
Furthermore, $y_1 < 0$ and $y_2 < 0$, because if $y_1 > 0$ we have $\langle x, y\rangle - f(x) = y_1 > 0$, for $x = e^1$, and same if $y_2 > 0$ we can choose $x = e^2.$
What bugs me is how do I prove that $D_{f^c} = \{y \in \mathbb{R^2} | y_1, y_2 < 0, y_1 y_2 \ge 1/4\}$. If I knew that I would have that $$x_1 (-y_1) + x_2(-y_2) = 2 \frac{x_1 (-y_1) + x_2(-y_2)}2 \ge 2 \sqrt{x_1 x_2 y_1 y_2} \ge \{ y_1 y_2 \ge \frac14\} \ge \sqrt{x_1 x_2}, $$ i.e. $ x_1 y_1 + x_2y_2 \le -\sqrt{x_1 x_2} \Leftrightarrow \langle x, y\rangle - f(x) \le 0.$ What I've noticed is that for $y_1 = y_2 = -x_1, x_1 = x_2$, I have $-2y_1^2 \le y_1,$ i.e. $y_1 \le -\frac12$ and same for $y_2 \le -\frac12$, so $y_1 y_2 \ge \frac14$, but this does not prove that the whole $D_{f^c} = \{y \in \mathbb{R^2} | y_1, y_2 < 0, y_1 y_2 \ge 1/4\}$, only that $\{y \in \mathbb{R^2} | y_1, y_2 < 0, y_1 y_2 \ge 1/4\} \subset D_{f^c}$.
Any help is appreciated.
What remains to check is the inclusion $$D_{f^c} \subset \{ y \in \mathbb R^2 \mid y_1,y_2 < 0, y_1y_2\ge1/4\}.$$ This is equivalent to $$\{ y \in \mathbb R^2 \mid y_1 \ge 0 \text{ or } y_2 \ge 0 \text{ or } y_1y_2 < 1/4\} \subset \mathbb R^2 \setminus (D_{f^c}).$$ Thus (by using your previous arguments), for $y$ in the set on the left-hand side you have to find $x$ with $\langle x, y\rangle - f(x) > 0$. Can you finish it from here?