Answer
$$\sqrt 3 + \frac{\pi }{6}$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}{\left. {\left( {x{{\sec }^{ - 1}}x} \right)} \right|_{x = \frac{2}{{\sqrt 3 }}}} \cr
& {\text{set }}y = x{\sec ^{ - 1}}x \cr
& {\text{differentiate }}y \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {x{{\sec }^{ - 1}}x} \right] \cr
& {\text{use product rule}} \cr
& \frac{{dy}}{{dx}} = x\frac{d}{{dx}}\left[ {{{\sec }^{ - 1}}x} \right] + {\sec ^{ - 1}}x\frac{d}{{dx}}\left[ x \right] \cr
& {\text{where }}\frac{d}{{dx}}\left[ {{{\sec }^{ - 1}}x} \right] = \frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }} \cr
& \frac{{dy}}{{dx}} = x\left( {\frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }}} \right) + {\sec ^{ - 1}}x \cr
& {\text{substitute }}y = x{\sec ^{ - 1}}x \cr
& \frac{d}{{dx}}\left( {x{{\sec }^{ - 1}}x} \right) = x\left( {\frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }}} \right) + {\sec ^{ - 1}}x \cr
& {\text{evaluate at }}x = \frac{2}{{\sqrt 3 }} \cr
& \frac{d}{{dx}}{\left. {\left( {x{{\sec }^{ - 1}}x} \right)} \right|_{x = \frac{2}{{\sqrt 3 }}}} = \left( {\frac{2}{{\sqrt 3 }}} \right)\left( {\frac{1}{{\left| {2/\sqrt 3 } \right|\sqrt {{{\left( {2/\sqrt 3 } \right)}^2} - 1} }}} \right) + {\sec ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right) \cr
& {\text{simplifying}} \cr
& \frac{d}{{dx}}{\left. {\left( {x{{\sec }^{ - 1}}x} \right)} \right|_{x = \frac{2}{{\sqrt 3 }}}} = \left( {\frac{1}{{\sqrt 3 /3}}} \right) + \frac{\pi }{6} \cr
& \frac{d}{{dx}}{\left. {\left( {x{{\sec }^{ - 1}}x} \right)} \right|_{x = \frac{2}{{\sqrt 3 }}}} = \sqrt 3 + \frac{\pi }{6} \cr} $$
