$\begin{aligned} & \text { Given, } x \sin (\alpha+y)=\sin y \text { and } y=\frac{m}{x^2+2 n x+1} \\ & x \sin (\alpha+y)=\sin y \\ & \Rightarrow \frac{\sin (\alpha+y)}{\sin y}=\frac{1}{x} \\ & \Rightarrow \frac{\sin \alpha \cos y+\cos \alpha \sin y}{\sin y}=\frac{1}{x} \\ & \Rightarrow \sin \alpha \cot y+\cos \alpha=\frac{1}{x} \\ & \Rightarrow \cot y=\frac{1-x \cos \alpha}{x \sin \alpha} \\ & \Rightarrow \tan y=\frac{x \sin \alpha}{1-x \cos \alpha} \\ & \Rightarrow \quad y=\tan { }^{-1}\left[\frac{x \sin \alpha}{1-x \cos \alpha}\right] \\ & \left.y^{\prime}=\frac{1}{1+\left(\frac{x \sin \alpha}{1-x \cos \alpha}\right)^2}\right] \\ & \left.\Rightarrow \quad y^{\prime}=\frac{[(1-x \cos \alpha) \sin \alpha-x \sin \alpha(-\cos \alpha)}{(1-x \cos \alpha)^2}\right] \\ & {\left[\frac{\sin \alpha-x \cos \alpha)^2+(x \sin \alpha)^2}{(1-x \cos \alpha)^2}\right.}\end{aligned}$
$$
\begin{aligned}
& \Rightarrow y^{\prime}=\frac{\sin \alpha}{1+x^2 \cos ^2 \alpha-2 x \cos \alpha+x^2 \sin ^2 \alpha} \\
& \Rightarrow y^{\prime}=\frac{\sin \alpha}{x^2-2 x \cos \alpha+1}
\end{aligned}
$$
Here, $m=\sin \alpha$ and $n=-\cos \alpha$
$$
\begin{aligned}
m^2 & =\sin ^2 \alpha \\
& =1-\cos ^2 \alpha=1-n^2
\end{aligned}
$$