L'Hopital's not Allowed

Relevant wiki: Maclaurin Series

We can use Maclaurin series to solve the problem as follows.

$\begin{aligned} L & = \lim_{x\to 0} \frac {x^2\sin x - x^3}{(e^x-1)(1-\cos(x^2))} & \small \color{#3D99F6} \text{Using Maclaurin series} \\ & = \lim_{x\to 0} \frac {x^2 \left(x-\frac {x^3}{3!} + \frac {x^5}{5!} - \cdots\right) - x^3}{\left(1+x+\frac {x^2}{2!} + \cdots -1\right)\left(1- \left(1-\frac {x^4}{2!} + \frac {x^8}{4!} - \cdots\right)\right)} \\ & = \lim_{x\to 0} \frac {-\frac {x^5}{3!} + \frac {x^7}{5!} - \frac {x^9}{7!} + \cdots}{\left(x+\frac {x^2}{2!} + \frac {x^3}{3!} \cdots\right)\left(\frac {x^4}{2!} - \frac {x^8}{4!} + \frac {x^{12}}{6!} \cdots \right)} \\ & = \lim_{x\to 0} \frac {-\frac {x^5}{3!} + \frac {x^7}{5!} - \frac {x^9}{7!} + O(x^{11})}{\frac {x^5}{2!} +\frac {x^6}{2!2!} + \frac {x^7}{3!2!} - O(x^9)} & \small \color{#3D99F6} \text{Divide up and down by }x^5 \\ & = \lim_{x\to 0} \frac {-\frac 1{3!} + \frac {x^2}{5!} - \frac {x^4}{7!} + O(x^6)}{\frac 1{2!} +\frac x{2!2!} + \frac {x^2}{3!2!} - O(x^4)} \\ & = - \frac {2!}{3!} = - \frac 13 \approx \boxed{-0.333} \end{aligned}$