Given, $x^2+y^2+\sin y=4$
After differentiating the above equation $w$.
r. t. $x$ we get
$$
\begin{aligned}
&2 x+2 y \frac{d y}{d x}+\cos y \frac{d y}{d x}=0 \ldots(1) \\
&\Rightarrow 2 x+(2 y+\cos y) \frac{d y}{d x}=0 \\
&\Rightarrow \frac{d y}{d x}=\frac{-2 x}{2 y+\cos y} \\
&\text { At }(-2,0),\left(\frac{d y}{d x}\right)_{(-2,0)}=\frac{-2 \times-2}{2 \times 0+\cos 0} \\
&\Rightarrow\left(\frac{d y}{d x}\right)_{(-2,0)}=\frac{4}{0+1} \\
&\Rightarrow\left(\frac{d y}{d x}\right)_{(2,0)}=4 \ldots(2)
\end{aligned}
$$
Again differentiating equation (1) w. r. t to $x$, we get
$2+2\left(\frac{d y}{d x}\right)^2+2 y \frac{d^2 y}{d x^2}-\sin y\left(\frac{d y}{d x}\right)^2+\cos \mathrm{y} \frac{d^2 y}{d x^2}=0$
$\Rightarrow 2+(-2 \sin y)\left(\frac{d y}{d x}\right)^2+(2 y+\cos \mathrm{y}) \frac{d^2 y}{d x^2}=0$
$\Rightarrow(2 y+\cos \mathrm{y}) \frac{d^2 y}{d x^2}=-2-(2-\sin y)\left(\frac{d y}{d x}\right)^2$
$\Rightarrow \frac{d^2 y}{d x^2}=\frac{-2-(2-\sin y)\left(\frac{d y}{d x}\right)^2}{2 y+\cos \mathrm{y}}$
So, at $(-2,0)$
$\frac{d^2 y}{d x^2}=\frac{-2-(2-0) \times 4^2}{2 \times 0+1}$
$\Rightarrow \frac{d^2 y}{d x^2}=\frac{-2-2 \times 16}{1}$
$\Rightarrow \frac{d^2 y}{d x^2}=-34$