It's just derivative of sine, right?

For $0 \le x \le \pi$ we have that $\sqrt{1 - \cos^{2}(x)}= \sin(x),$ and thus the right-sided limit is

$\large \lim_{x \rightarrow 0^{+}} \left( \dfrac{d}{dx}\sqrt{1 - \cos^{2}(x)} \right) = \lim_{x \rightarrow 0^{+}} \left(\dfrac{d}{dx} \sin(x) \right) = \lim_{x \rightarrow 0^{+}} \cos(x) = 1.$

For $-\pi \le x \le 0$ we have that $\sqrt{1 - \cos^{2}(x)} = -\sin(x),$ and thus the left-sided limit is

$\large \lim_{x \rightarrow 0^{-}} \left( \dfrac{d}{dx} (-\sin(x)) \right) = \lim_{x \rightarrow 0^{-}} (-\cos(x)) = -1.$

Since the right-sided and left-sided limits are different, the given two-sided limit does not exist, i.e., the correct answer is $\boxed{DNE}.$