$\begin{array}{l} \left\{ \begin{array}{l} \sin x + \cos x = \dfrac{1}{3}\\ {\sin ^2}x + {\cos ^2}x = 1 \end{array} \right. \Rightarrow {\left( {\sin x + \cos x} \right)^2} = \dfrac{1}{9}\\ \Rightarrow {\sin ^2}x + {\cos ^2}x + 2\sin x\cos x = \dfrac{1}{9}\\ \Rightarrow 2\sin x\cos x = \dfrac{{ - 8}}{9} \Rightarrow 4\sin x\cos x = \dfrac{{ - 16}}{9}\\ + {\left( {\sin x - \cos x} \right)^2} = {\left( {\sin x + \cos x} \right)^2} - 4\sin x\cos x = \dfrac{1}{3} + \dfrac{{16}}{9} = \dfrac{{19}}{9}\\ \Rightarrow \left| {\sin x - \cos x} \right| = \dfrac{{\sqrt {19} }}{3} \end{array}$
