Answer
${f^{\left( 3 \right)}}\left( 0 \right) = - 8$
Work Step by Step
We compute the derivatives of $f\left( x \right) = \dfrac{1}{{1 + \tan x}}$ and list them in the table below:
$\begin{array}{*{20}{c}}
n&{\frac{{{f^{\left( n \right)}}\left( x \right)}}{{n!}}}&{}&{}&{}&{}&{}&{}&{}&{\frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}}\\
0&{\frac{1}{{1 + \tan \left( x \right)}}}&{}&{}&{}&{}&{}&{}&{}&1\\
1&{ - \frac{1}{{{{\left( {\sin \left( x \right) + \cos \left( x \right)} \right)}^2}}}}&{}&{}&{}&{}&{}&{}&{}&{ - 1}\\
2&{\frac{{\cos \left( x \right) - \sin \left( x \right)}}{{{{\left( {\sin \left( x \right) + \cos \left( x \right)} \right)}^3}}}}&{}&{}&{}&{}&{}&{}&{}&1\\
3&{\frac{{2\left( {\sin \left( {2x} \right) - 2} \right)}}{{3{{\left( {\sin \left( x \right) + \cos \left( x \right)} \right)}^4}}}}&{}&{}&{}&{}&{}&{}&{}&{ - \frac{4}{3}}\\
4&{ - \frac{{11\sin \left( x \right) + \sin \left( {3x} \right) - 11\cos \left( x \right) + \cos \left( {3x} \right)}}{{6{{\left( {\sin \left( x \right) + \cos \left( x \right)} \right)}^5}}}}&{}&{}&{}&{}&{}&{}&{}&{\frac{5}{3}}\\
5&{\frac{{26\sin \left( {2x} \right) + \cos \left( {4x} \right) - 33}}{{15{{\left( {\sin \left( x \right) + \cos \left( x \right)} \right)}^6}}}}&{}&{}&{}&{}&{}&{}&{}&{ - \frac{{32}}{{15}}}
\end{array}$
According to the Maclaurin series: $f\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n}$.
From the table above, we get the coefficient of ${x^3}$, which is $ - \dfrac{4}{3}$.
So, $\dfrac{{{f^{\left( 3 \right)}}\left( 0 \right)}}{{3!}} = - \dfrac{4}{3}$.
Therefore, ${f^{\left( 3 \right)}}\left( 0 \right) = - 8$.
