1. **State the problem:** Find the 9th term of a geometric sequence where the first term $a_1=5$ and the common ratio $r=\frac{1}{3}$.\n\n2. **Formula for the nth term of a geometric sequence:** $$a_n = a_1 \times r^{n-1}$$\nThis formula means to find the nth term, multiply the first term by the common ratio raised to the power of $n-1$.\n\n3. **Apply the formula:** For the 9th term, $n=9$, so\n$$a_9 = 5 \times \left(\frac{1}{3}\right)^{9-1} = 5 \times \left(\frac{1}{3}\right)^8$$\n\n4. **Calculate the power:**\n$$\left(\frac{1}{3}\right)^8 = \frac{1^8}{3^8} = \frac{1}{6561}$$\n\n5. **Multiply:**\n$$a_9 = 5 \times \frac{1}{6561} = \frac{5}{6561}$$\n\n6. **Final answer:** The 9th term of the sequence is $$\boxed{\frac{5}{6561}}$$.