Tell whether $\dfrac{10^{91}-1}{9}$ is prime or not? I really have no idea how to start. The only theorem considering prime numbers I know of is Fermat's little theorem and maybe its related with binomial theorem.

Any help will be appreciated.
Just think through the actual number.

$10^{91}$ is a $1$ with $91$ $0$'s after it.

$10^{91}-1$ is therefore $91$ $9$'s in a row.

$\frac{10^{91}-1}{9}$ is therefore $91$ $1$'s in a row.

Due to the form of this number, $x$ $1$'s in a row will divide it, where $x$ is a divisor of $91$.

For example $1111111$ is a divisor, so is $1111111111111$.

Hence the number is not prime.
You have

$$111111....1=1+10+10^2.....+10^{90}=(10^{84}+10^{77}+10^70.....+10^7+1) (10^6+10^5+10^4+ \dots + 10^0);$$
this is product of two integers 2 or greater,  so it is not prime.