Tenemos la indeterminación de tipo
0/0,
tal que el límite para el numerador es
$$\lim_{x \to 0^+} \tan^{2}{\left(3 x \right)} = 0$$
y el límite para el denominador es
$$\lim_{x \to 0^+} \frac{1}{\cot^{2}{\left(x \right)}} = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to 0^+}\left(\tan^{2}{\left(3 x \right)} \cot^{2}{\left(x \right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \tan^{2}{\left(3 x \right)}}{\frac{d}{d x} \frac{1}{\cot^{2}{\left(x \right)}}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\left(6 \tan^{2}{\left(3 x \right)} + 6\right) \tan{\left(3 x \right)} \cot^{3}{\left(x \right)}}{2 \cot^{2}{\left(x \right)} + 2}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{6 \tan{\left(3 x \right)} \cot^{3}{\left(x \right)}}{2 \cot^{2}{\left(x \right)} + 2}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \frac{1}{2 \cot^{2}{\left(x \right)} + 2}}{\frac{d}{d x} \frac{1}{6 \tan{\left(3 x \right)} \cot^{3}{\left(x \right)}}}\right)$$
=
$$\lim_{x \to 0^+}\left(- \frac{2 \left(- 2 \cot^{2}{\left(x \right)} - 2\right) \cot{\left(x \right)}}{\left(\frac{- 3 \tan^{2}{\left(3 x \right)} - 3}{6 \tan^{2}{\left(3 x \right)} \cot^{3}{\left(x \right)}} + \frac{3 \cot^{2}{\left(x \right)} + 3}{6 \tan{\left(3 x \right)} \cot^{4}{\left(x \right)}}\right) \left(2 \cot^{2}{\left(x \right)} + 2\right)^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(- \frac{2 \left(- 2 \cot^{2}{\left(x \right)} - 2\right) \cot{\left(x \right)}}{\left(\frac{- 3 \tan^{2}{\left(3 x \right)} - 3}{6 \tan^{2}{\left(3 x \right)} \cot^{3}{\left(x \right)}} + \frac{3 \cot^{2}{\left(x \right)} + 3}{6 \tan{\left(3 x \right)} \cot^{4}{\left(x \right)}}\right) \left(2 \cot^{2}{\left(x \right)} + 2\right)^{2}}\right)$$
=
$$9$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 2 vez (veces)