1. **State the problem:** Solve the equation $$(3x - 1)^{-\frac{2}{3}} = \frac{1}{4} \cdot \frac{3}{2}.$$\n\n2. **Simplify the right side:** Multiply the fractions:\n$$\frac{1}{4} \cdot \frac{3}{2} = \frac{3}{8}.$$\nSo the equation becomes:\n$$(3x - 1)^{-\frac{2}{3}} = \frac{3}{8}.$$\n\n3. **Rewrite the equation:** Recall that $a^{-b} = \frac{1}{a^b}$, so\n$$(3x - 1)^{-\frac{2}{3}} = \frac{1}{(3x - 1)^{\frac{2}{3}}} = \frac{3}{8}.$$\n\n4. **Invert both sides:** To isolate $(3x - 1)^{\frac{2}{3}}$, take the reciprocal of both sides:\n$$\frac{1}{(3x - 1)^{\frac{2}{3}}} = \frac{3}{8} \implies (3x - 1)^{\frac{2}{3}} = \frac{8}{3}.$$\n\n5. **Raise both sides to the power $\frac{3}{2}$ to solve for $3x - 1$:**\n$$\left((3x - 1)^{\frac{2}{3}}\right)^{\frac{3}{2}} = \left(\frac{8}{3}\right)^{\frac{3}{2}} \implies 3x - 1 = \left(\frac{8}{3}\right)^{\frac{3}{2}}.$$\n\n6. **Calculate $\left(\frac{8}{3}\right)^{\frac{3}{2}}$:**\nFirst, write as\n$$\left(\frac{8}{3}\right)^{\frac{3}{2}} = \left(\sqrt{\frac{8}{3}}\right)^3 = \left(\frac{\sqrt{8}}{\sqrt{3}}\right)^3 = \left(\frac{2\sqrt{2}}{\sqrt{3}}\right)^3.$$\n\n7. **Simplify inside the cube:**\n$$\left(\frac{2\sqrt{2}}{\sqrt{3}}\right)^3 = \frac{(2)^3 (\sqrt{2})^3}{(\sqrt{3})^3} = \frac{8 \cdot 2^{\frac{3}{2}}}{3^{\frac{3}{2}}} = \frac{8 \cdot 2 \sqrt{2}}{3 \sqrt{3}} = \frac{16 \sqrt{2}}{3 \sqrt{3}}.$$\n\n8. **Rationalize the denominator:**\n$$\frac{16 \sqrt{2}}{3 \sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{16 \sqrt{6}}{9}.$$\n\nSo,\n$$3x - 1 = \frac{16 \sqrt{6}}{9}.$$\n\n9. **Solve for $x$:**\n$$3x = 1 + \frac{16 \sqrt{6}}{9} = \frac{9}{9} + \frac{16 \sqrt{6}}{9} = \frac{9 + 16 \sqrt{6}}{9}.$$\n\nDivide both sides by 3:\n$$x = \frac{9 + 16 \sqrt{6}}{9 \cdot 3} = \frac{9 + 16 \sqrt{6}}{27}.$$\n\n**Final answer:**\n$$\boxed{x = \frac{9 + 16 \sqrt{6}}{27}}.$$