We have indeterminateness of type
0/0,
i.e. limit for the numerator is
$$\lim_{x \to 0^+}\left(- x + \operatorname{atan}{\left(x \right)}\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} x^{3} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{- x + \operatorname{atan}{\left(x \right)}}{x^{3}}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to 0^+}\left(\frac{- x + \operatorname{atan}{\left(x \right)}}{x^{3}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- x + \operatorname{atan}{\left(x \right)}\right)}{\frac{d}{d x} x^{3}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{-1 + \frac{1}{x^{2} + 1}}{3 x^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(-1 + \frac{1}{x^{2} + 1}\right)}{\frac{d}{d x} 3 x^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(- \frac{1}{3 \left(x^{2} + 1\right)^{2}}\right)$$
=
$$\lim_{x \to 0^+} - \frac{1}{3}$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- 2 x\right)}{\frac{d}{d x} 6 x}\right)$$
=
$$\lim_{x \to 0^+} - \frac{1}{3}$$
=
$$\lim_{x \to 0^+} - \frac{1}{3}$$
=
$$- \frac{1}{3}$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 3 time(s)