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!! 

$$\begin{align*}\lim_{x\to y} \frac{\sin^2(x)-\sin^2(y)}{x^2-y^2}&=\lim_{x\to y}\frac{\sin x-\sin y}{x-y}\frac{\sin x+\sin y}{x+y}\\&=\frac{2\sin y}{2y}\lim_{x\to y}\frac{\sin x-\sin y}{x-y}\\&=\frac{\sin y}{y}(\sin y)'\\&=\frac{\sin y \cos y}{y}\end{align*}$$
$$=\frac{\sin2y}{2y}$$
