Answer
The error in the expression is $\sqrt{-9}\cdot \sqrt{-9}\ne \sqrt{81}$ but the expression ${{\left( \sqrt{-9} \right)}^{2}}=\sqrt{-9}\cdot \sqrt{-9}$ is correct.
Work Step by Step
Consider the expression, ${{\left( \sqrt{-9} \right)}^{2}}=\sqrt{-9}\cdot \sqrt{-9}=\sqrt{81}$
Take the first part of the expression.
${{\left( \sqrt{-9} \right)}^{2}}=\sqrt{-9}\cdot \sqrt{-9}$
The principal square root of a negative number is such that for any positive real number $b$,
$\sqrt{-b}=i\sqrt{b}$
Express all the square roots of a negative numbers in terms of $i$.
$\begin{align}
& {{\left( \sqrt{-9} \right)}^{2}}=\left( i\sqrt{9} \right)\cdot \left( i\sqrt{9} \right) \\
& ={{i}^{2}}{{\left( \sqrt{9} \right)}^{2}}
\end{align}$
The imaginary unit is $i=\sqrt{-1}$, where ${{i}^{2}}=-1$.
Replace the value ${{i}^{2}}=-1$.
$\begin{align}
& {{\left( \sqrt{-9} \right)}^{2}}=\left( -1 \right){{\left( \sqrt{9} \right)}^{2}} \\
& =\left( -1 \right)\left( 9 \right) \\
& =-9
\end{align}$
The value of ${{\left( \sqrt{-9} \right)}^{2}}$ is equal to $-9$.
The error in the expression is such that, ${{\left( \sqrt{-9} \right)}^{2}}=\sqrt{-9}\cdot \sqrt{-9}$ and $\sqrt{-9}\cdot \sqrt{-9}\ne \sqrt{81}$
$\begin{align}
& \sqrt{-9}\cdot \sqrt{-9}\ne \sqrt{\left( -9 \right)\left( -9 \right)} \\
& \ne \sqrt{{{\left( -1 \right)}^{2}}{{\left( 9 \right)}^{2}}} \\
& \ne \sqrt{81}
\end{align}$
Therefore, the error in the expression is $\sqrt{-9}\cdot \sqrt{-9}\ne \sqrt{81}$ but the expression ${{\left( \sqrt{-9} \right)}^{2}}=\sqrt{-9}\cdot \sqrt{-9}$ is correct.
