How do you simplify ${{\left( 6{{y}^{2}} \right)}^{-2}}$?

Hint: We first explain the process of exponents and indices. We find the general form. Then we explain the different binary operations on exponents. We use the identities to find the simplified form of ${{\left( 6{{y}^{2}} \right)}^{-2}}$ with positive exponents.

Complete step by step solution:
We know the exponent form of the number $a$ with the exponent being $n$ can be expressed as ${{a}^{n}}$. In case the value of $n$ becomes negative, the value of the exponent takes its inverse value. The formula to express the form is ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}},n\in {{\mathbb{R}}^{+}}$.
If we take two exponential expressions where the exponents are $m$ and $n$.
Let the numbers be ${{a}^{m}}$ and ${{a}^{n}}$. We take multiplication of these numbers.
The indices get added. So, ${{a}^{m+n}}={{a}^{m}}\times {{a}^{n}}$.
The division works in an almost similar way. The indices get subtracted. So, $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$.
We also have the identity of ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ and ${{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}$.
For a given equation ${{\left( 6{{y}^{2}} \right)}^{-2}}$, we break the constant and the variable part separately.
Applying ${{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}$, we get ${{\left( 6{{y}^{2}} \right)}^{-2}}={{6}^{-2}}{{\left( {{y}^{2}} \right)}^{-2}}$.
Applying ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}},n\in {{\mathbb{R}}^{+}}$ for the constant we get ${{6}^{-2}}=\dfrac{1}{{{6}^{2}}}=\dfrac{1}{36}$.
Now for ${{\left( {{y}^{2}} \right)}^{-2}}$, using ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ we get ${{\left( {{y}^{2}} \right)}^{-2}}={{y}^{2\times \left( -2 \right)}}={{y}^{-4}}=\dfrac{1}{{{y}^{4}}}$
So, final form be ${{\left( 6{{y}^{2}} \right)}^{-2}}=\dfrac{1}{36{{y}^{4}}}$
Therefore, the simplified form of ${{\left( 6{{y}^{2}} \right)}^{-2}}$ using only positive exponents is $\dfrac{1}{36{{y}^{4}}}$.

Note: The addition and subtraction for exponents works for taking common terms out depending on the values of the indices.
For numbers ${{a}^{m}}$ and ${{a}^{n}}$, we have ${{a}^{m}}\pm {{a}^{n}}={{a}^{m}}\left( 1\pm {{a}^{n-m}} \right)$.the relation is independent of the values of $m$ and $n$. We need to remember that the condition for ${{a}^{m}}={{a}^{n}}\Rightarrow m=n$ is that the value of $a\ne 0,\pm 1$.