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