1. **State the problem:** Solve the equation $$\left(\frac{3}{7}\right)^{3x-7} = \left(\frac{7}{3}\right)^{7x-3}.$$\n\n2. **Rewrite the right side:** Note that $$\frac{7}{3} = \left(\frac{3}{7}\right)^{-1}.$$ So the equation becomes $$\left(\frac{3}{7}\right)^{3x-7} = \left(\left(\frac{3}{7}\right)^{-1}\right)^{7x-3} = \left(\frac{3}{7}\right)^{-(7x-3)}.$$\n\n3. **Set the exponents equal:** Since the bases are the same and nonzero, the exponents must be equal: $$3x - 7 = -(7x - 3).$$\n\n4. **Simplify the equation:** $$3x - 7 = -7x + 3.$$\n\n5. **Bring all terms to one side:** $$3x + 7x = 3 + 7,$$ which simplifies to $$10x = 10.$$\n\n6. **Solve for x:** $$x = \frac{10}{10} = 1.$$\n\n**Final answer:** $$\boxed{1}.$$