Evaluating $\displaystyle\lim_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x^2-y^2}$ 
I have to evaluate $\displaystyle\lim_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x^2-y^2}$

I tried several ways like replacing $x^2-y^2$ by $z$ and then solving and breaking the denominator into $(x+y)(x-y)$ and then replacing $(x-y)$ by $z$ and then solving but always I end up with complicated things which are nowhere near the answer.
How do I solve?
Thanks for any help!! 

$$\lim _{ x\rightarrow y } \frac { \sin ^{ 2 } x-\sin ^{ 2 } y }{ x^{ 2 }-y^{ 2 } } =\lim _{ x\rightarrow y } \frac { \left( \sin { x-\sin { y }  }  \right) \left( \sin { x+\sin { y }  }  \right)  }{ \left( x-y \right) \left( x+y \right)  } =\\ =\lim _{ x\rightarrow y } \frac { \left( 2\sin { \frac { x-y }{ 2 }  } \cos { \frac { x+y }{ 2 }  }  \right) \left( \sin { x+\sin { y }  }  \right)  }{ \left( x-y \right) \left( x+y \right)  } =\\ =\lim _{ x\rightarrow y } \frac { \sin { \frac { x-y }{ 2 }  }  }{ \frac { \left( x-y \right)  }{ 2 }  } \frac { \cos { \frac { x+y }{ 2 }  } \left( \sin { x+\sin { y }  }  \right)  }{ \left( x+y \right)  } =\\ =\cos { y } \frac { 2\sin { y }  }{ 2y } =\frac { \sin { y\cos { y }  }  }{ y } $$
