1. Simplify the algebraic part

Given

$$y=\frac{(\sqrt{x}+1)(x^2-\sqrt{x})}{x\sqrt{x}+x+\sqrt{x}}+\frac1{15}(3\cos^2 x-5)\cos^3 x$$

Let $t=\sqrt{x}$. Then $x=t^2$ and

$$x^2-\sqrt{x}=t^4-t=t(t^3-1)=t(t-1)(t^2+t+1)$$

and

$$(\sqrt{x}+1)=t+1.$$

So the numerator of the first fraction is

$$(t+1)\,t(t-1)(t^2+t+1)=t(t^2-1)(t^2+t+1).$$

The denominator is

$$x\sqrt{x}+x+\sqrt{x}=t^3+t^2+t=t(t^2+t+1).$$

Hence the first term simplifies to

$$\frac{t(t^2-1)(t^2+t+1)}{t(t^2+t+1)}=t^2-1=x-1.$$

Therefore,

$$y=x-1+\frac1{15}(3\cos^2 x-5)\cos^3 x.$$

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2. Differentiate

Differentiate term by term:

$$y'=1+\frac{d}{dx}\left[\frac1{15}(3\cos^2 x-5)\cos^3 x\right].$$

First simplify the trigonometric expression:

$$\frac1{15}(3\cos^2 x-5)\cos^3 x n=\frac1{15}(3\cos^5 x-5\cos^3 x).$$

So

$$y'=1+\frac1{15}\frac{d}{dx}(3\cos^5 x-5\cos^3 x).$$

Now,

$$\frac{d}{dx}(\cos^5 x)=5\cos^4 x(-\sin x),$$ $$\frac{d}{dx}(\cos^3 x)=3\cos^2 x(-\sin x).$$

Thus

$$\frac{d}{dx}(3\cos^5 x-5\cos^3 x) =3\cdot 5\cos^4 x(-\sin x)-5\cdot 3\cos^2 x(-\sin x).$$ $$= -15\cos^4 x\sin x+15\cos^2 x\sin x$$ $$=15\sin x\cos^2 x(1-\cos^2 x) =15\sin x\cos^2 x\sin^2 x.$$

Therefore,

$$y'=1+\sin^3 x\cos^2 x.$$

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3. Evaluate at $x=\dfrac{\pi}{6}$

We know

$$\sin\frac{\pi}{6}=\frac12, \qquad \cos\frac{\pi}{6}=\frac{\sqrt3}{2}.$$

So

$$\sin^3\frac{\pi}{6}\cos^2\frac{\pi}{6} =\left(\frac12\right)^3\left(\frac{\sqrt3}{2}\right)^2 =\frac18\cdot\frac34 =\frac{3}{32}.$$

Hence

$$y'\left(\frac{\pi}{6}\right)=1+\frac{3}{32}=\frac{35}{32}.$$

Now,

$$96\,y'\left(\frac{\pi}{6}\right)=96\cdot\frac{35}{32}=3\cdot 35=105.$$

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4. Comparison with stored answer

Derived answer is

$$105$$

which matches the stored correct answer.