imc

Prime factorizing $201^9$, we get $3^9\cdot 67^9$. A perfect square must have even powers of its prime factors, so our possible choices for our exponents to get perfect square are $0,2,4,6,8$ for both $3$ and $67$. This yields $5\cdot 5=25$ perfect squares. Perfect cubes must have multiples of $3$ for each of their prime factors' exponents, so we have either $0,3,6$, or $9$ for both $3$ and $67$, which yields $4\cdot 4=16$ perfect cubes, for a total of $25+16=41$. Subtracting the overcounted powers of $6$ ($3^0\cdot 67^0$ , $3^0\cdot 67^6$ , $3^6\cdot 67^0$, and $3^6\cdot 67^6$), we get $41-4=\boxed{37}$.