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!! 

$\lim_\limits{x\to y} \frac{\sin^2 x - \sin^2 y}{x^2-y^2} =\lim_\limits{x\to y} \left(\frac{\sin x - \sin y}{x-y}\right)\left(\frac{\sin x + \sin y}{x+y}\right)\\
(\frac {d}{dy} \sin y)(\frac {\sin y}{y})$
