How do you simplify \[{{\left( 3m \right)}^{-2}}\] and write it using only positive exponents?

Hint: In the given question, we have been asked to simplify an expression. In order to simplify the expression, we need to remove any grouping symbol such as brackets by multiplying the given factor. Then use any exponent rule to remove grouping if the terms within the bracket are containing exponent. Combine the like terms using mathematical operations such as addition and subtraction. Final answer should always in positive exponent, if there is any negative exponent use the definition of negative exponent i.e. \[{{a}^{-n}}=\dfrac{1}{{{a}^{n}}}\] to make the final answer in positive exponent.

Formula used:
If \[n\] is a positive integer and \[a\] is not equal to \[0\], then
\[{{a}^{-n}}=\dfrac{1}{{{a}^{n}}}\].

Complete step by step answer:
We have the given expression:
\[{{\left( 3m \right)}^{-2}}\]
Using the definition of a negative exponent i.e. \[{{a}^{-n}}=\dfrac{1}{{{a}^{n}}}\], we get
\[{{\left( 3m \right)}^{-2}}=\dfrac{1}{{{\left( 3m \right)}^{2}}}\]
By using exponent property i.e. \[{{\left( ab \right)}^{m}}={{a}^{m}}\times {{b}^{m}}\], we get
\[{{\left( 3m \right)}^{-2}}=\dfrac{1}{{{3}^{2}}\times {{m}^{2}}}\]
Taking the square in right-hand side of the equation, we get
\[\therefore{{\left( 3m \right)}^{-2}}=\dfrac{1}{9{{m}^{2}}}\]

Therefore, \[\dfrac{1}{9{{m}^{2}}}\] is the required solution.

Note:If \[n\] is a positive integer and \[a\]is not equal to \[0\], then \[{{a}^{-n}}=\dfrac{1}{{{a}^{n}}}\]. The negative exponent in any given expression tells us to rewrite the same expression by doing the reciprocal of the base with the changing sign of the exponent. If there is any expression that has a negative exponent, then it will not be called the simplest form. To answer any question, we will use the concept of negative exponent and the other properties of exponent to write the given expression with only positive exponent.