$\log \sqrt[3]{3/6}=\log {{(3/6)}^{\frac{1}{3}}}=\frac{1}{3}\log (\frac{36}{10})=\frac{1}{3}(\log 36-\log 10)=\frac{1}{3}(\log ({{6}^{2}})-1)$ $=\frac{1}{3}(2\log (2\times 3)-1)=\frac{1}{3}(2(\log 2+\log 3)-1)=\frac{1}{3}(2(\underbrace{0/3+0/5}_{0/8})-1)=\frac{1}{3}(1/6-1)=\frac{0/6}{3}=0/2$