PLEASE NOTE : I have no idea if this is correct - someone please check
sin(x+y)=cosy
differentiate y in terms of x
$$\begin{array}{rll}
sin(x+y)&=&cosy\\\\
cos(x+y)[1+\frac{dy}{dx}]&=&-(siny)\frac{dy}{dx}\\\\
cos(x+y)+cos(x+y)\frac{dy}{dx}+(siny)\frac{dy}{dx}&=&0\\\\
cos(x+y)\frac{dy}{dx}+(siny)\frac{dy}{dx}&=&-cos(x+y)\\\\
\frac{dy}{dx}[cos(x+y)+(siny)]&=&-cos(x+y)\\\\
\frac{dy}{dx}&=&\frac{-cos(x+y)}{cos(x+y)+(siny)}\\\\
\end{array}$$
If that is correct then i guess
$$\frac{dx}{dy}&=&\frac{cos(x+y)+(siny)}{-cos(x+y)}\\\\$$
PLEASE NOTE : I have no idea if this is correct - someone please check
sin(x+y)=cosy
differentiate y in terms of x
$$\begin{array}{rll}
sin(x+y)&=&cosy\\\\
cos(x+y)[1+\frac{dy}{dx}]&=&-(siny)\frac{dy}{dx}\\\\
cos(x+y)+cos(x+y)\frac{dy}{dx}+(siny)\frac{dy}{dx}&=&0\\\\
cos(x+y)\frac{dy}{dx}+(siny)\frac{dy}{dx}&=&-cos(x+y)\\\\
\frac{dy}{dx}[cos(x+y)+(siny)]&=&-cos(x+y)\\\\
\frac{dy}{dx}&=&\frac{-cos(x+y)}{cos(x+y)+(siny)}\\\\
\end{array}$$
If that is correct then i guess
$$\frac{dx}{dy}&=&\frac{cos(x+y)+(siny)}{-cos(x+y)}\\\\$$