If x^2+y^2+ y=4 , then the value of (d^2 y)/(d x^2) at the point...

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