Think outside the box (2)

Coefficient of $x^n$ in $f(x)$ is $\displaystyle \frac{1}{n!}f^{(n)}(0)$ .

so using simple methods we determine coefficient of $x^3$ is $-99\times 100^{98}$ .

$\displaystyle \frac{1}{3!}f^{(3)}(0)=-99\times 100^{98}\implies f^{(3)}(0)=\boxed{-594}\times 100^{98}$

Let $y(x) = 9x^2 - x^3 - 100$ , then we have:

\(\begin{array} {} y (x) = 9x^2 - x^3 - 100 & \implies y(0) = - 100 \\ y' (x) = 18x - 3x^2 & \implies y'(0) = 0 \\ y'' (x) = 18 - 6x & \implies y''(0) = 0 \\ y''' (x) = - 6x & \implies y'''(0) = -6 \\ y^{(n)} (x) = 0 \text{ for } n \ge 4 \end{array} \)

$\begin{aligned} f^{(0)}(x) & = y^{99} \\ f^{(1)}(x) & = 99y^{98}y' \\ f^{(2)}(x) & = 99 \cdot 98 y^{97}(y')^2 + 99y^{98}y'' \\ f^{(3)}(x) & = 99 \cdot 98 \cdot 97 y^{96}(y')^3 + 99 \cdot 98 y^{97}\cdot 2y'y'' + 99 \cdot 98 y^{97}y'y'' + 99y^{98}y''' \\ f^{(3)}(0) & = 0 + 0 + 0 + 99(-100)^{98}(-6) \\ & = -594 \cdot 100^{98} \end{aligned}$

$\implies k = \boxed{594}$