1. Given differential equation

$\sqrt{1-x^2}\,\frac{dy}{dx}+\sqrt{1-y^2}=0$

So,

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

This is separable.

2. Separate variables

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

Integrating both sides,

$\int \frac{dy}{\sqrt{1-y^2}}= -\int \frac{dx}{\sqrt{1-x^2}}$

Using

$\int \frac{dt}{\sqrt{1-t^2}}=\sin^{-1} t,$

we get

$\sin^{-1} y=-\sin^{-1} x + C$

or

$\sin^{-1} y + \sin^{-1} x = C$

3. Use the initial condition

Given

$y\left(\frac12\right)=\frac{\sqrt3}{2}$

Substitute:

$\sin^{-1}\left(\frac{\sqrt3}{2}\right)+\sin^{-1}\left(\frac12\right)=C$

Now,

$\sin^{-1}\left(\frac{\sqrt3}{2}\right)=\frac{\pi}{3}, \qquad \sin^{-1}\left(\frac12\right)=\frac{\pi}{6}$

Hence,

$C=\frac{\pi}{3}+\frac{\pi}{6}=\frac{\pi}{2}$

So the solution satisfies

$\sin^{-1} y + \sin^{-1} x = \frac{\pi}{2}$

Thus,

$\sin^{-1} y = \frac{\pi}{2} - \sin^{-1} x$

Taking sine on both sides,

$y=\sin\left(\frac{\pi}{2}-\sin^{-1}x\right)=\cos(\sin^{-1}x)=\sqrt{1-x^2}$

Since $|x|<1$ and principal values are used, this is valid here.

4. Find $y\left(-\frac{1}{\sqrt2}\right)$

$y\left(-\frac{1}{\sqrt2}\right)=\sqrt{1-\left(-\frac{1}{\sqrt2}\right)^2}$

$=\sqrt{1-\frac12}$

$=\sqrt{\frac12}=\frac{1}{\sqrt2}$

5. Check options

The value is

$\frac{1}{\sqrt2}$

So the correct option is C.