From Förberedande kurs i matematik 1

That which is under the root sign is the same as \displaystyle \left( -\text{3} \right)^{\text{2}}=\text{9 } and because \displaystyle \text{9}=\text{3}\centerdot \text{3}=\text{3}^{\text{2}}, hence

\displaystyle \sqrt{\left( -3 \right)^{2}}=\sqrt{9}=9^{{1}/{2}\;}=\left( 3^{2} \right)^{{1}/{2}\;}=3^{2\centerdot \frac{1}{2}}=3^{1}=3

NOTE: The calculation \displaystyle \sqrt{\left( -3 \right)^{2}}=\left( \left( -3 \right)^{2} \right)^{{1}/{2}\;}=\left( -3 \right)^{2\centerdot \frac{1}{2}}=\left( -3 \right)^{1}=-3

is wrong at the second equals sign. Remember that the power rules apply when the base is positive.