Two methods are giving two different answers to the this differential equation : $\frac{dy}{dx}=\frac{1}{2} \frac{d(\sin ^{-1}(f(x))}{dx}$

There is no way to predict the value of $y(-\sqrt{3})$ by knowing the solution over $(1,\infty)$.

The problem is ill posed. It is like asking for the value of $y(-1)$ if $y'=1/x$ for $x\ne0$, with $y(1)=0$. Any value for $y(-1)$ can be chosen.

Indeed, there is no reason for the constant of integration to be the same over $(0,\infty)$ and $(-\infty,0)$ in this case or over $(1,\infty)$ and $(-\infty,-1)$ in your case.

By the way, we have $$ f(x)=\begin{cases} \frac{\pi}{2}-\arctan x & x>1 \\[6px] -\frac{\pi}{2}-\arctan x & x<-1 \end{cases} $$ as witnessed by https://www.desmos.com/calculator/6akvm0e78h

Therefore the differential equation is $$ y'=-\frac{1}{1+x^2} $$ and so $$ y=\begin{cases} a-\arctan x & x>1 \\[6px] b-\arctan x & x<-1 \end{cases} $$ You can determine $a$ by plugging in $a-\arctan\sqrt{3}=\pi/6$, so $a=\pi/2$. However, this does not impose any condition on $b$.

If the instructor wants you to use $b=a$, then the answer would be $$ \frac{\pi}{2}-\arctan(-\sqrt{3})=\frac{\pi}{2}+\frac{\pi}{3}=\frac{5\pi}{6} $$ but there is no mathematical justification for this and the instructor is wrong.