Given,

$\sec y\frac{dy}{dx}-\sin \left(x+y\right)-\sin \left(x-y\right)=0$

$\Rightarrow \sec y\frac{dy}{dx}=2\sin x\cos y$

$\Rightarrow {\sec }^{2}ydy=2\sin xdx$

$\Rightarrow \int {\sec }^{2}ydy=\int 2\sin xdx$

$\Rightarrow \tan y=-2\cos x+c$

Given, $y\left(0\right)=0$

So, $0=-2\cos 0+c$

$\Rightarrow c=2$

So, the particular solution is 

$\tan y=-2\cos x+2$

at $x=\frac{\pi }{2},\tan y=-2\cos \frac{\pi }{2}+2$

$\Rightarrow \tan y=2\Rightarrow {\sec }^{2}y=5$

Here, differentiate both sides of $\tan y=-2\cos x+2,$ we get,

${\sec }^{2}y\frac{dy}{dx}=2\sin x$

At $x=\frac{\mathrm{\pi }}{2}$

$5\frac{dy}{dx}=2\sin \left(\frac{\pi }{2}\right)$

$\Rightarrow 5\frac{dy}{dx}=2$