Answer
$$ y = \ln \left| {\frac{{x + \sqrt {{x^2} - 9} }}{3}} \right|$$
Work Step by Step
$$\eqalign{
& \sqrt {{x^2} - 9} \frac{{dy}}{{dx}} = 1,\,\,\,\,\,\,\,x > 3,\,\,\,\,\,\,y\left( 5 \right) = \ln 3 \cr
& {\text{Separate variables}} \cr
& dy = \frac{1}{{\sqrt {{x^2} - 9} }}dx \cr
& {\text{then}} \cr
& y = \int {\frac{1}{{\sqrt {{x^2} - 9} }}} dx \cr
& {\text{We set}}{\text{, }}x = 3\sec \theta,{\text{ }}dx = 3\sec \theta \tan \theta d\theta \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& y = \int {\frac{1}{{\sqrt {9{{\sec }^2}\theta - 9} }}} \left( {3\sec \theta \tan \theta d\theta } \right) \cr
& y = \int {\frac{1}{{\sqrt {9\left( {{{\sec }^2}\theta - 1} \right)} }}} \left( {3\sec \theta \tan \theta d\theta } \right) \cr
& {\text{use the fundamental identity }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr
& y = \int {\frac{1}{{\sqrt {9{{\tan }^2}\theta } }}} \left( {3\sec \theta \tan \theta d\theta } \right) \cr
& {\text{simplifying}} \cr
& y = \int {\frac{1}{{3\tan \theta }}} \left( {3\sec \theta \tan \theta d\theta } \right) \cr
& y = \int {\sec \theta } d\theta \cr
& {\text{integrate}} \cr
& y = \ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& {\text{write in terms of }}x, \cr
& y = \ln \left| {\frac{x}{3} + \frac{{\sqrt {{x^2} - 9} }}{3}} \right| + C \cr
& {\text{use initial condition }}y\left( 5 \right) = \ln 3 \cr
& \ln 3 = \ln \left| {\frac{5}{3} + \frac{{\sqrt {{5^2} - 9} }}{3}} \right| + C \cr
& \ln 3 = \ln \left| {\frac{5}{3} + \frac{4}{3}} \right| + C \cr
& \ln 3 = \ln \left| 3 \right| + C \cr
& C = 0 \cr
& {\text{then}} \cr
& y = \ln \left| {\frac{x}{3} + \frac{{\sqrt {{x^2} - 9} }}{3}} \right| \cr
& y = \ln \left| {\frac{{x + \sqrt {{x^2} - 9} }}{3}} \right| \cr} $$
