Question:
Suppose $m$ is a two-digit positive integer such that $6^{-1}\pmod m$ exists and $6^{-1}\equiv 6^2\pmod m$. What is $m$?

Answer:
We can multiply both sides of the congruence $6^{-1}\equiv 6^2\pmod m$ by $6$: $$
\underbrace{6\cdot 6^{-1}}_1 \equiv \underbrace{6\cdot 6^2}_{6^3} \pmod m.
$$Thus $6^3-1=215$ is a multiple of $m$. We know that $m$ has two digits. The only two-digit positive divisor of $215$ is $43$, so $m=\boxed{43}$.