If $x\operatorname{Sin}\left( a+y \right)+\operatorname{Sin}a\operatorname{Cos}\left( a+y \right)=0$ , prove that $\dfrac{dy}{dx}=\dfrac{{{\operatorname{Sin}}^{2}}\left( a+y \right)}{\operatorname{Sin}a}$ .

Hint: Separate the terms involving $x$ and $y$ , put arrangement involving $x$ in Right Hand Side and arrangement involving $y$ in Left Hand Side and then differentiate both side with respect to $x$ using the chain rule $\dfrac{d}{dx}f\left( g\left( x \right) \right)=f'\left( g\left( x \right) \right)g'\left( x \right)$ and hence solve for $\dfrac{dy}{dx}$ to get the desired equality.

Complete step by step answer:
We are given an implicit equation involving variables $x$ and $y$ ,
And we have to prove a term for $\dfrac{dy}{dx}$
First consider the given implicit equation ,
 $x\operatorname{Sin}\left( a+y \right)+\operatorname{Sin}a\operatorname{Cos}\left( a+y \right)=0$
Now, we will separate the terms for $x$ and $y$ ,
$\begin{align}
  & x\operatorname{Sin}\left( a+y \right)=-\operatorname{Sin}a\operatorname{Cos}\left( a+y \right) \\
 & \dfrac{\operatorname{Sin}\left( a+y \right)}{\operatorname{Cos}\left( a+y \right)}=-\dfrac{\operatorname{Sin}a}{x}
\end{align}$
Since, we need reciprocal of $\operatorname{Sin}a$ and the term in the proof does not involve $x$ so , we will reciprocate on both the sides and get,
$\operatorname{Cot}\left( a+y \right)=-\dfrac{x}{\operatorname{Sin}a}$
Since , The Left Hand Side involves function of a function we have to use the chain rule here , i.e.
$\dfrac{d}{dx}f\left( g\left( x \right) \right)=f'\left( g\left( x \right) \right)g'\left( x \right)$
Now, differentiating both sides with respect to $''x''$ , we get,
(For differentiating the Left Hand Side , we will use, $\dfrac{d}{dx}\operatorname{Cot}x=-{{\operatorname{Cosec}}^{2}}x$ )
$\begin{align}
  & -{{\operatorname{Cosec}}^{2}}\left( a+y \right)\dfrac{dy}{dx}=-\dfrac{1}{\operatorname{Sin}a} \\
 & \dfrac{dy}{dx}=\dfrac{1}{{{\operatorname{Cosec}}^{2}}\left( a+y \right)\operatorname{Sin}a} \\
 & \dfrac{dy}{dx}=\dfrac{{{\operatorname{Sin}}^{2}}\left( a+y \right)}{\operatorname{Sin}a}
\end{align}$ Since, $\dfrac{1}{\operatorname{Cosec}x}=\operatorname{Sin}x$

Hence, $\dfrac{dy}{dx}=\dfrac{{{\operatorname{Sin}}^{2}}\left( a+y \right)}{\operatorname{Sin}a}$

Note: At first when we see the the question , there is the possibility that someone try to differentiate the implicit equation itself with respect to $''x''$ , without doing any changes in the equation and then solve for $\dfrac{dy}{dx}$ . This is not wrong but this could complicate the equation for solving for $\dfrac{dy}{dx}$ , and it may happen that we could not even prove the equality . So, the best approach is to first separate it into terms , first involving $''x''$ and second involving $''y''$ and then differentiating both sides with respect to $''x''$ and get the desired results.