By definition of the Sine Integral:
$\displaystyle \displaystyle \dfrac{d}{dx}\left[\text{Si}(x^6)\right] = \dfrac{d}{dx}\int_0^x \dfrac{\sin \left(t^6\right)}{t^6}dt$
By the Fundamental Theorem of Calculus (sometimes called Part I):
$\displaystyle \displaystyle \dfrac{d}{dx}\int_0^x \dfrac{\sin \left(t^6\right)}{t^6}dt = \dfrac{\sin(x^6)}{x^6}$
This is simply a direct application of the theorem. There is no calculation necessary. Just apply the Fundamental Theorem of Calculus and the answer is immediately apparent.
Ok, I am not familiar with the Si function. I looked it up, but apparently do not understand it. According to Wolframalpha,
$\displaystyle \dfrac{d}{dx}\left[\text{Si}(x^6)\right] = \dfrac{6\sin(x^6)}{x}$
But, I have no clue how.
Oh, I just misread the description:
$\displaystyle \displaystyle \text{Si}(x^6) = \int_0^{x^6}\dfrac{\sin t}{t} dt$
So, by the Chain Rule, we have:
$\displaystyle \dfrac{d}{dx}\left[\text{Si}(x^6)\right] = \dfrac{d}{d(x^6)}\left[\text{Si}(x^6)\right]\dfrac{d(x^6)}{dx} = \dfrac{\sin(x^6)}{x^{\cancel{6}}}(6\cancel{x^5}) = \dfrac{6\sin(x^6)}{x}$