Given equation

$x^2 + y^2 + \,\sin y = 4$

We need to find:

$\frac{d^2 y}{dx^2}$

at the point $(-2,0)$.

Differentiate implicitly once

Differentiate both sides with respect to $x$:

$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) + \frac{d}{dx}(\sin y) = 0$

$2x + 2y\frac{dy}{dx} + \cos y\frac{dy}{dx} = 0$

So,

$2x + (2y + \cos y)y' = 0$

where $y' = \frac{dy}{dx}$.

Hence,

$y' = \frac{-2x}{2y + \cos y}$

Find $y'$ at $(-2,0)$

Substitute $x=-2, y=0$:

$y' = \frac{-2(-2)}{2(0) + \cos 0} = \frac{4}{1} = 4$

Differentiate again

From

$2x + (2y + \cos y)y' = 0$

differentiate with respect to $x$:

$2 + \frac{d}{dx}\left[(2y + \cos y)y'\right] = 0$

Use product rule:

$2 + \left(\frac{d}{dx}(2y + \cos y)\right)y' + (2y + \cos y)y'' = 0$

Now,

$\frac{d}{dx}(2y + \cos y) = 2y' - \sin y\, y' = (2 - \sin y)y'$

Therefore,

$2 + (2 - \sin y)(y')^2 + (2y + \cos y)y'' = 0$

So,

$y'' = \frac{-2 - (2 - \sin y)(y')^2}{2y + \cos y}$

Evaluate at $(-2,0)$

At $y=0$,

$\sin 0 = 0, \quad \cos 0 = 1, \quad y' = 4$

Hence,

$y'' = \frac{-2 - (2-0)(4)^2}{2(0)+1}$

$y'' = -2 - 2\cdot 16$

$y'' = -2 - 32 = -34$

Compare with options

The correct option is:

$\boxed{\text{A: } -34}$

Comparison with stored answer

Stored correct answer is A, which matches our result.