Suppose $A=\displaystyle \frac{dy}{dx}$ when $x^2+y^2=4$ at $(\sqrt{2},\sqrt{2})$,$B=\displaystyle \frac{dy}{dx}$ when $\sin y+ \sin x=\sin x-\sin y$ at $(\pi,\pi)$ and $C=\displaystyle \frac{dy}{dx}$ when $2e^{xy}+e^x e^y-e^x-e^y=e^{xy+1}$ at $(1,1)$, then $(A+B+C)$ has the value equal to

$A: \displaystyle \frac {d}{dx} (x^2+y^2=4)$ at $(\sqrt 2, \sqrt 2))$

$2x+2y\displaystyle \frac { dy }{ dx } =0$

$\displaystyle \frac { dy }{ dx } =-\displaystyle \frac { x }{ y } =-\displaystyle \frac { \sqrt { 2 } }{ \sqrt { 2 } } =-1$

$B: \displaystyle \frac {d}{dx}(\sin y+\sin x=\sin x\cdot \sin y)$ at $(\pi, \pi)$

$\cos { y\displaystyle \frac { dy }{ dx } } +\cos { x } =\sin { x } \cos { y\displaystyle \frac { dy }{ dx } +\sin { y } \cos { x } }$

$\displaystyle \frac { dy }{ dx } \left( \cos { y } -\sin { x } \cos { y } \right) =\sin { y\cos { x-\cos { x } } }$

$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { \cos { x\left( \sin { y-1 } \right) } }{ \cos { y\left( 1-\sin { x } \right) } }$

$B=\displaystyle \frac { -1\left( 0-1 \right) }{ -1\left( 1-0 \right) } =-1$

$C: \displaystyle \frac {d}{dx}(2e^{xy}+e^xe^y-e^x-{ e }^{ y }=e^{xy+1})$ at $(1, 1)$

$\left[ 2{ e }^{ xy }\left( x\displaystyle \frac { dy }{ dx } +y \right) \right] +{ e }^{ x }{ e }^{ y }\displaystyle \frac { dy }{ dx } +{ e }^{ y }{ e }^{ x }-{ e }^{ x } -{ e }^{ y }\frac { dy }{ dx } ={ e }^{ xy+1 }\left( x\displaystyle \frac { dy }{ dx } +y \right)$

$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { y{ e }^{ xy+1 }-2y{ e }^{ xy }-{ e }^{ x+y }+{ e }^{ x } }{ 2x{ e }^{ xy }+{ e }^{ x+y }-x{ e }^{ xy+1 }-{ e }^{ y } }$..........

since ${ e }^{ x }{ e }^{ y }={ e }^{ x+y }$

$C=\displaystyle \frac { { e }^{ 2 }-2{ e }^{ 1 }-{ e }^{ 2 }+{ e }^{ 1 } }{ 2{ e }^{ 1 }+{ e }^{ 2 }-{ e }^{ 2 }-{ e }^{ 1 } }=-{1}$

$A+B+C=-1-1-1=-3$