On the probability of getting the same number for three dice

The probability that the first die is a 1 is $\frac16$. The probability that the second die is a 1 is $\frac16$. The probability that the third die is a 1 is $\frac16$. Multiplying this together gives $\frac1{6^3}$ for the probability that all dice are a 1.

We have the same chance $\left(\frac1{6^3}\right)$ of rolling three 2s, three 3s, three 4s, three 5s, or three 6s, so the probability of rolling any number on all three dice at the same time is $6\cdot\frac{1}{6^3}$, or $\frac1{36}$.

Also note that both your answer $\frac{1}{6^2}$ and $6\cdot\frac{1}{6^3}$ are equal to $\frac1{36}$.

By counting: There are $6^3$ possible outcomes: 1-1-1, 1-1-2, ..., 6-6-5, 6-6-6. There are $6$ outcomes with all the numbers the same: 1-1-1, 2-2-2, ..., 5-5-5, 6-6-6. So the probability of all the numbers being the same is $6/6^3$.

Alternatively: Throw the first die. It doesn't matter what number you get.

Throw the second die. The probability of getting the same number is $1/6$.

Throw the third die. The probability of getting the same number is again $1/6$.

So the probability of three numbers the same is $1/6 \times 1/6$.

Just a small supplement to the nice answers addressing your question, since I didn't found it mentioned explicitly.

We can calculate the probability of an event by calculating the ratio of the number of favorable choices or number of successes of an event with the number of all choices of an event.

This way we obtain \begin{align*} \frac{\text{number of successes}}{\text{number of all choices}}=\frac{6}{6^3}=\frac{1}{6^2}=\frac{1}{36} \end{align*}