6. Let $A, B$ be points on the graph of the function $f(x)=3-x^{2}$ on either side of the $y$-axis, then the minimum value of the area of the region enclosed by the tangent lines of $f(x)$ at points $A, B$ and the $x$-axis is $\qquad$ .
Given $A\left(a, 3-a^{2}\right), B\left(-b, 3-b^{2}\right)$,
where $a, b>0$, then the tangent line $A D$:
$$
\begin{array}{l}
y=-2 a(x-a)+3-a^{2} \\
\Rightarrow y=-2 a x+a^{2}+3,
\end{array}
$$

Similarly, we get $B E: y=2 b x+b^{2}+3$.
Thus, $D\left(\frac{a^{2}+3}{2 a}, 0\right), E\left(-\frac{b^{2}+3}{2 b}, 0\right)$,
and by solving the system, we get $C\left(\frac{a-b}{2}, a b+3\right)$
$$
\begin{array}{l}
\Rightarrow S_{\triangle C D E}=\frac{1}{2}\left(\frac{a^{2}+3}{2 a}+\frac{b^{2}+3}{2 b}\right)(a b+3)=\frac{1}{4}\left(a+\frac{3}{a}+b+\frac{3}{b}\right)(a b+3) \\
=\frac{1}{4}(a+b)\left(1+\frac{3}{a b}\right)(a b+3) \geqslant \frac{1}{2} \cdot \sqrt{a b}\left(a b+\frac{9}{a b}+6\right) \xlongequal{t=\sqrt{a b>0}} \frac{1}{2} t\left(t^{2}+\frac{9}{t^{2}}+6\right)=\frac{1}{2}\left(t^{3}+\frac{9}{t}+6 t\right),
\end{array}
$$

Let $f(t)=t^{3}+\frac{9}{t}+6 t, f^{\prime}(t)=3 t^{2}-\frac{9}{t^{2}}+6=\frac{\left(t^{2}-1\right)\left(t^{2}+3\right)}{t^{2}}$, then $f(t)$ is decreasing on $(0,1)$ and increasing on $(1,+\infty)$, so $f(t) \geqslant f(1)=16 \Rightarrow S_{\triangle C D E} \geqslant 8$.
Therefore, the minimum area of the region is 8.