Answer
$$dy = \frac{{2{{\sec }^2}x\left[ {\left( {{x^2} + 1} \right)\tan x - x} \right]}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx$$
Work Step by Step
$$\eqalign{
& y = \frac{{{{\sec }^2}x}}{{{x^2} + 1}} \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \frac{{dy}}{{dx}} = \underbrace {\frac{d}{{dx}}\left[ {\frac{{{{\sec }^2}x}}{{{x^2} + 1}}} \right]}_{{\text{Quotient rule}}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {{x^2} + 1} \right)\left( {{{\sec }^2}x} \right)' - {{\sec }^2}x\left( {{x^2} + 1} \right)'}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {{x^2} + 1} \right)\left( {2\sec x} \right)\left( {\sec x\tan } \right) - {{\sec }^2}x\left( {2x} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{2\left( {{x^2} + 1} \right){{\sec }^2}x\tan x - 2x{{\sec }^2}x}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr
& {\text{Factoring the numerator}} \cr
& \frac{{dy}}{{dx}} = \frac{{2{{\sec }^2}x\left[ {\left( {{x^2} + 1} \right)\tan x - x} \right]}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr
& {\text{Write in differential form }}dy = f'\left( x \right)dx \cr
& dy = \frac{{2{{\sec }^2}x\left[ {\left( {{x^2} + 1} \right)\tan x - x} \right]}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx \cr} $$
