Why multiplying powers of prime factors of a number yields number of total divisors?
Solution 1:
If $d$ divides $36$, then no prime numbers other than $2$ and $3$ can divide $d$. On the other hand, $36=2^23^2$ and so $d=2^\alpha3^\beta$, with $\alpha,\beta\in\{0,1,2\}$. Since there are three possibilities for $\alpha$ and another $3$ for $\beta$, there are $9(=3\times3)$ possibilities for $d$.
Solution 2:
For 36 all the divisors are of the form $2^s3^k$, where $0 \le s,k \le 2$. Thus as you have 3 choices for each exponent the number of divisors is $3 \cdot 3 = 9$.