Answer
${f^{\left( 3 \right)}}\left( 0 \right) = - 6$
Work Step by Step
From Table 2 of Section 11.7, we have
${{\rm{e}}^x} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n!}} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$.
Substituting ${x^2}$ for $x$ in the series above we get
${{\rm{e}}^{{x^2}}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^{2n}}}}{{n!}}$, ${\ \ \ }$ converges for all $x$
Thus,
$f\left( x \right) = \left( {{x^2} - x} \right){{\rm{e}}^{{x^2}}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{\left( {{x^2} - x} \right){x^{2n}}}}{{n!}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^{2n + 2}} - {x^{2n + 1}}}}{{n!}}$
Write out the first three terms:
$f\left( x \right) = \left( {{x^2} - x} \right){{\rm{e}}^{{x^2}}} = \left( {{x^2} - x} \right) + \left( {{x^4} - {x^3}} \right) + \dfrac{1}{2}\left( {{x^6} - {x^5}} \right) + \cdot\cdot\cdot$
$f\left( x \right) = \left( {{x^2} - x} \right){{\rm{e}}^{{x^2}}} = - x + {x^2} - {x^3} + {x^4} - \dfrac{1}{2}{x^5} + \dfrac{1}{2}{x^6} + \cdot\cdot\cdot$
Notice that the coefficient of ${x^3}$ is $-1$.
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
$\dfrac{{{f^{\left( 3 \right)}}\left( 0 \right)}}{{3!}} = - 1$
So, ${f^{\left( 3 \right)}}\left( 0 \right) = - 6$
