Answer
$$z = \frac{1}{{6 - {{\tan }^{ - 1}}x}}$$
Work Step by Step
$$\eqalign{
& \frac{{dz}}{{dx}} = \frac{{{z^2}}}{{1 + {x^2}}} \cr
& {\text{separate}} \cr
& \frac{{dz}}{{{z^2}}} = \frac{{dx}}{{1 + {x^2}}} \cr
& {\text{integrate both sides}} \cr
& \int {\frac{{dz}}{{{z^2}}}} = \int {\frac{{dx}}{{1 + {x^2}}}} \cr
& - \frac{1}{z} = {\tan ^{ - 1}}x + C \cr
& {\text{the initial condition }}z\left( 0 \right) = \frac{1}{6}{\text{ implies that}} \cr
& - \frac{1}{{1/6}} = {\tan ^{ - 1}}\left( 0 \right) + C \cr
& C = - 6 \cr
& - \frac{1}{z} = {\tan ^{ - 1}}x - 6 \cr
& z = \frac{1}{{6 - {{\tan }^{ - 1}}x}} \cr} $$
