Answer
$$y' = \frac{1}{{\sqrt {9 + {x^2}} }}$$
Work Step by Step
$$\eqalign{
& y = {\sinh ^{ - 1}}\left( {\frac{1}{3}x} \right) \cr
& {\text{find the derivatvive}} \cr
& y' = \left( {{{\sinh }^{ - 1}}\left( {\frac{1}{3}x} \right)} \right)' \cr
& {\text{use the theorem 6}}{\text{.8}}{\text{.5 }}\left( {{\text{ see page 479}}} \right){\text{ }} \cr
& \frac{d}{{dx}}\left[ {{{\sinh }^{ - 1}}u} \right] = \frac{1}{{\sqrt {1 + {u^2}} }}\frac{{du}}{{dx}} \cr
& y' = \frac{1}{{\sqrt {1 + {{\left( {x/3} \right)}^2}} }}\left( {\frac{x}{3}} \right)' \cr
& y' = \frac{1}{{\sqrt {1 + {{\left( {x/3} \right)}^2}} }}\left( {\frac{1}{3}} \right) \cr
& {\text{simplifying}} \cr
& y' = \frac{1}{{3\sqrt {\frac{{9 + {x^2}}}{9}} }} \cr
& y' = \frac{1}{{3\left( {\frac{{\sqrt {9 + {x^2}} }}{3}} \right)}} \cr
& y' = \frac{1}{{\sqrt {9 + {x^2}} }} \cr} $$
