The use of $a^{-1}$ instead of $\dfrac 1a$ (or $\div a$) is often but not always required. Basically, it's complicated.
With a little numerical manipulation, both ways will make sense; but, you still need to be careful.
For example, we find $3^{-1} \equiv 9 \pmod{13}$ (Note that $3 \times 9 \equiv 27 \equiv 1 \pmod{13}$). In the following cases, I will show that some times multiplying by $3^{-1} \equiv 9 \pmod{13}$ is appropriate, and some times dividing by $3$ is appropriate.
(1.) Consider the equivalence, $3x \equiv 12 \pmod{13}.$
So we can compute $ x \equiv 9 \times 12 \equiv 108 \equiv 4 \pmod{13}$. But it would be silly to do all of that work when we can see that $12 \div 3 = 4$.
(2.) Consider the equivalence, $3x \equiv 10 \pmod{13}.$
Since $10 \div 3$ is not an integer; we can't use that method. By repeatedly adding $13$, we see that $10 \equiv 23 \equiv 36 \pmod{13}$ and $36 \div 3 =12.$ So $x \equiv 12 \pmod{13}$. This method is only practical though for very small moduli. It would be much more efficient, usually, to just compute $x = 3^{-1}\times 10 \equiv 9 \times 10 \equiv 90 \equiv 12 \pmod{13}$.
There is a theorem that states that the equivalence $ax \equiv b \pmod c$ if and only if $\gcd(a,c) \mid b$. If there is a solution, then the equivalence reduce to $\dfrac ag x \equiv \dfrac bg \pmod{\dfrac cg}$ where $g = \gcd(a,c)$.
(3.) Consider the equivalence, $3x \equiv 12 \pmod{15}.$
In this case, $\gcd(3,15)=3 \mid 12$, so there is a solution and we find $x \equiv 4 \pmod{5}.$
(4.) Consider the equivalence, $3x \equiv 10 \pmod{15}.$
In this case, $\gcd(3,15)=3 \not \mid 10$, so there is no solution.