6. The set of all real number pairs $(x, y)$ that satisfy $\left\{\begin{array}{l}4^{-x}+27^{-y}=\frac{5}{6} \\ \log _{27} y-\log _{4} x \geqslant \frac{1}{6} \text { is } \\ 27^{y}-4^{x} \leqslant 1\end{array}\right.$ $\qquad$ . $\qquad$
6. $\left\{\left(\frac{1}{2}, \frac{1}{3}\right)\right\}$ Detailed Explanation: Let $a=4^{x}, b=27^{y}$, then $\frac{1}{a}+\frac{1}{b}=\frac{5}{6}, a>\frac{6}{5}, b>\frac{6}{5}$. From $\frac{1}{a}+\frac{1}{b}=\frac{5}{6}$, we get $b=\frac{6 a}{5 a-6}$, then $\frac{6 a}{5 a-6}-a \leqslant 1$, solving this yields $a \geqslant 2$ or $a>\frac{6}{5}$, so $a \geqslant 2$. Similarly, we get $b \leqslant 3$. Therefore, $x \geqslant \frac{1}{2}, y \leqslant \frac{1}{3} . \log _{27} y-\log _{4} x \leqslant \log _{27} \frac{1}{3}-\log _{4} \frac{1}{2}=\frac{1}{6}$. So $x=\frac{1}{2}, y=\frac{1}{3}$.