It is given
$$\lim_{x \to 0} \frac{d}{dx}\,\frac{\sin^2 x}{x} = \lim_{x \to 0} \frac{d}{dx} \sin^2(x)x^{-1}$$
Using chain rule, we get
$$\lim_{x \to 0} [-1.\sin^2(x)x^{-2} + \sin(2x)x^{-1}]$$
$$-1.\Big(\underbrace{\lim_{x \to 0} \frac{\sin(x)}{x}}_{1}\Big).\Big(\underbrace{\lim_{x \to 0} \frac{\sin(x)}{x}}_{1}\Big) + 2.\Big(\underbrace{\lim_{ x \to 0} \frac{\sin(2x)}{2x}}_{1}\Big)$$
This gives, $-1.1.1 + 2.1 = 2-1=1$
$\textbf{Option (C) is correct}$