Which is the fastest way to check if a number has only 3 unique factors ? Any help will be highly appreciated?
The fastest method will depend on the types of numbers you intend to check and how you are running the tests (pencil and paper, x86_64, etc.). But basically, check if the number has any small prime factors (say, the first 25: 2, 3, 5, ..., 97). If so, the number has exactly 3 divisors if and only if the number is the square of that prime. Otherwise you have a number with no tiny prime factors; now check if the number is a square. You can first test mod some promising moduli, like 63, 64, and 65. If it passes that test, take the integer square root $s$ and test if that, squared, gives the number. If not, the number has more than 3 divisors. If so, test if the remaining number is prime. The first step is using a probable-prime test, maybe checking if the number is a b-strong pseudoprime for some randomly-chosen $1 < b < s-1.$ If that fails the number is composite and so $s^2$ has more than three divisors. Otherwise apply a primality-proving test to $s$.
Supposing that by "only 3 divisors" you mean "exactly 3 divisors", the only such numbers are numbers of the form $p^2$, where $p$ is prime; they are divisible by $1, p, $ and $p^2$. So the first thing to try is to test if the number $n$ is a perfect square, which you can do efficiently by calculating the integer square root $i = \lfloor \sqrt n\rfloor$ and then checking to see if $i^2=n$. If not, $n$ fails. If so, $n$ has the desired property if and only if $i$ is prime, which you can check using any of the usual primality tests.