We have indeterminateness of type
0/0,
i.e. limit for the numerator is
$$\lim_{x \to 0^+} \sin^{2}{\left(x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \tan{\left(x^{2} \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin^{2}{\left(x \right)}}{\tan{\left(x^{2} \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin^{2}{\left(x \right)}}{\frac{d}{d x} \tan{\left(x^{2} \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(x \right)} \cos{\left(x \right)}}{x \left(\tan^{2}{\left(x^{2} \right)} + 1\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(x \right)}}{x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} 2 \sin{\left(x \right)}}{\frac{d}{d x} 2 x}\right)$$
=
$$\lim_{x \to 0^+} \cos{\left(x \right)}$$
=
$$\lim_{x \to 0^+} 1$$
=
$$\lim_{x \to 0^+} 1$$
=
$$1$$
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)