Answer
First method
$$A^{-2}=(A^2)^{-1}=\left[ \begin {array}{cc} \frac{11}{4}&\frac{3}{2}\\ \frac{3}{4}&\frac{1}{2}
\end {array} \right] .
$$
Second method
$$A^{-2}=(A^{-1})^{2}=\left[ \begin {array}{cc} \frac{11}{4}&\frac{3}{2}\\ \frac{3}{4}&\frac{1}{2}
\end {array} \right] .$$
One can see that the two methods are equivalent.
Work Step by Step
First method; we find $A^2$ which is given by
$$A^{2}=\left[ \begin {array}{cc} 2&-6\\ -3&11\end {array}
\right]
].$$
By calculating the inverse of $A^2$, we have
$$A^{-2}=(A^2)^{-1}=\left[ \begin {array}{cc} \frac{11}{4}&\frac{3}{2}\\ \frac{3}{4}&\frac{1}{2}
\end {array} \right] .
$$
Second method; we find $A^{-1}$ as follows
$$A^{-1}=\left[ \begin {array}{cc} -\frac{3}{2}&-1\\ -\frac{1}{2}&0
\end {array} \right] .
$$
Now, we have
$$A^{-2}=(A^{-1})^{2}=\left[ \begin {array}{cc} \frac{11}{4}&\frac{3}{2}\\ \frac{3}{4}&\frac{1}{2}
\end {array} \right] .$$
One can see that the two methods are equivalent.