Answer
$$f'\left( x \right) = \frac{{6x\tan x - 2x - 3\left( {{x^2} + 1} \right){{\sec }^2}x}}{{{{\left( {3\tan x - 1} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{\left( {{x^2} + 1} \right)\cot x}}{{3 - \cos x\csc x}} \cr
& {\text{Where }}\csc x = \frac{1}{{\sin x}} \Rightarrow \cos x\left( {\frac{1}{{\sin x}}} \right) = \cot x \cr
& f\left( x \right) = \frac{{\left( {{x^2} + 1} \right)\cot x}}{{3 - \cot x}} \cr
& {\text{Multiply by }}\frac{{\tan x}}{{\tan x}} \cr
& f\left( x \right) = \frac{{{x^2} + 1}}{{3\tan x - 1}} \cr
& \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{{\left( {3\tan x - 1} \right)\frac{d}{{dx}}\left[ {{x^2} + 1} \right] - \left( {{x^2} + 1} \right)\frac{d}{{dx}}\left[ {3\tan x - 1} \right]}}{{{{\left( {3\tan x - 1} \right)}^2}}} \cr
& {\text{Computing derivatives}} \cr
& f'\left( x \right) = \frac{{\left( {3\tan x - 1} \right)\left( {2x} \right) - \left( {{x^2} + 1} \right)\left( {3{{\sec }^2}x} \right)}}{{{{\left( {3\tan x - 1} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{6x\tan x - 2x - 3\left( {{x^2} + 1} \right){{\sec }^2}x}}{{{{\left( {3\tan x - 1} \right)}^2}}} \cr} $$
