Answer
${f^{\left( 3 \right)}}\left( 0 \right) = 2$
Work Step by Step
From Table 2 of Section 11.7:
${\tan ^{ - 1}}x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{2n + 1}} = x - \dfrac{{{x^3}}}{3} + \dfrac{{{x^5}}}{5} - \dfrac{{{x^7}}}{7} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for $\left| x \right| \le 1$.
Substituting $\left( {{x^2} - x} \right)$ for $x$ in the series above gives
${\tan ^{ - 1}}\left( {{x^2} - x} \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{{\left( {{x^2} - x} \right)}^{2n + 1}}}}{{2n + 1}}$
Write out the first three terms:
${\tan ^{ - 1}}\left( {{x^2} - x} \right) = \left( {{x^2} - x} \right) - \dfrac{{{{\left( {{x^2} - x} \right)}^3}}}{3} + \dfrac{{{{\left( {{x^2} - x} \right)}^5}}}{5} + \cdot\cdot\cdot$
$ = \left( {{x^2} - x} \right) - \dfrac{{ - {x^3} + 3{x^4} - 3{x^5} + {x^6}}}{3} + \dfrac{{ - {x^5} + 5{x^6} - 10{x^7} + 10{x^8} - 5{x^9} + {x^{10}}}}{5} + \cdot\cdot\cdot$
Notice that the coefficient of ${x^3}$ is $\dfrac{1}{3}$.
According to the Maclaurin series: $f\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n}$.
Comparing with the expansion above, we get
$\dfrac{{{f^{\left( 3 \right)}}\left( 0 \right)}}{{3!}} = \dfrac{1}{3}$
So, ${f^{\left( 3 \right)}}\left( 0 \right) = 2$
