Divide numerator and denominator of the first argument by $\cos x$:

$\frac{3\cos x-4\sin x}{4\cos x+3\sin x}=\frac{\frac{3}{4}-\tan x}{1+\frac{3}{4}\tan x}.$

This is the tangent subtraction formula, so the first term equals ${\tan }^{-1}\left(\frac{3}{4}\right)-{\tan }^{-1}(\tan x)$, which on a neighbourhood of $x=\frac{\sqrt{3}}{2}$ simplifies to ${\tan }^{-1}\left(\frac{3}{4}\right)-x$.

For the second term, substitute $x=\sin \theta$:

$2{\tan }^{-1}\left(\frac{\sin \theta }{1+\cos \theta }\right)=2\cdot \frac{\theta }{2}=\theta ={\sin }^{-1}x.$

So $y={\tan }^{-1}\left(\frac{3}{4}\right)-x+{\sin }^{-1}x$, giving

$\frac{dy}{dx}=-1+\frac{1}{\sqrt{1-x^2}}.$

At $x=\frac{\sqrt{3}}{2}$: $\sqrt{1-3/4}=\frac{1}{2}$, so $\frac{dy}{dx}=-1+2=1.$