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