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