9. In a right-angled tetrahedron $A B C D$, $D A, D B, D C$ are mutually perpendicular. Let $S, R$ represent its surface area and the radius of the circumscribed sphere, respectively. Then the maximum value of $\frac{S}{R^{2}}$ is $\qquad$.
9. $\frac{2}{3}(3+\sqrt{3})$ Let $D A=a, D B=b, D C=c$,

then $2 S=a b+b c+c a+\sqrt{a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}}$
$$
\leqslant a^{2}+b^{2}+c^{2}+\frac{\sqrt{3}}{3}\left(a^{2}+b^{2}+c^{2}\right)=\frac{4(3+\sqrt{3})}{3} R^{2}
$$

Thus $\frac{S}{R^{2}} \leqslant \frac{2(3+\sqrt{3})}{3}$,
with equality holding if and only if $a=b=c$.