Answer
$$y = 10\operatorname{arcsec} \left| x \right| - 10\operatorname{arcsec} \left( 3 \right)$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{{10}}{{x\sqrt {{x^2} - 1} }} \cr
& {\text{Separate the variables}} \cr
& dy = \frac{{10}}{{x\sqrt {{x^2} - 1} }}dx \cr
& {\text{Integrate both sides}} \cr
& \int {dy} = \int {\frac{{10}}{{x\sqrt {{x^2} - 1} }}} dx \cr
& y = 10\int {\frac{1}{{x\sqrt {{x^2} - 1} }}} dx \cr
& {\text{By Theorem 5}}{\text{.17 }}\int {\frac{{du}}{{u\sqrt {{u^2} - {a^2}} }} = \frac{1}{a}\operatorname{arcsec} \frac{{\left| u \right|}}{a} + C} \cr
& y = 10\left( {\frac{1}{1}\operatorname{arcsec} \frac{{\left| x \right|}}{1}} \right) + C \cr
& y = 10\operatorname{arcsec} \left| x \right| + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}y\left( 3 \right) = 0 \cr
& 0 = 10\operatorname{arcsec} \left| 3 \right| + C \cr
& C = - 10\operatorname{arcsec} \left( 3 \right) \cr
& {\text{Substituting }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = 10\operatorname{arcsec} \left| x \right| - 10\operatorname{arcsec} \left( 3 \right) \cr} $$
