Answer
$$\frac{\pi }{{12}}$$
Work Step by Step
$$\eqalign{
& \int_{\sqrt 2 /3}^{2/3} {\frac{{dy}}{{\left| y \right|\sqrt {9{y^2} - 1} }}} \cr
& {\text{Use the substitution method:}} \cr
& u = 3y,\,\,\,\,\,\,du = 3dy,\,\,\,\,dy = \frac{{du}}{3} \cr
& {\text{the new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}y = 2/3,{\text{ then }}u = 2 \cr
& \,\,\,\,\,\,{\text{If }}y = \sqrt 2 /3,{\text{ then }}u = \sqrt 2 \cr
& {\text{then}} \cr
& \int_{\sqrt 2 /3}^{2/3} {\frac{{dy}}{{\left| y \right|\sqrt {9{y^2} - 1} }}} = \int_{\sqrt 2 }^2 {\frac{{du/3}}{{\left| y \right|\sqrt {{u^2} - 1} }}} \cr
& = \int_{\sqrt 2 }^2 {\frac{{du}}{{\left| u \right|\sqrt {{u^2} - 1} }}} \cr
& {\text{integrate}} \cr
& = \left( {{{\sec }^{ - 1}}u} \right)_{\sqrt 2 }^2 \cr
& {\text{use the fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = {\sec ^{ - 1}}\left( 2 \right) - {\sec ^{ - 1}}\left( {\sqrt 2 } \right) \cr
& {\text{simplifying, we get:}} \cr
& = \frac{\pi }{3} - \frac{\pi }{4} \cr
& = \frac{\pi }{{12}} \cr} $$
