If  $x=\sec \theta -\cos \theta$  and  $y={\sec }^n\theta -{\cos }^n\theta$  then

$\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{n{\sec }^n\theta Tan\theta {\cos }^{n-1}\theta \sin \theta }{\sec \theta Tan\theta +\sin \theta }=\frac{n({\sec }^n\theta +{\cos }^n\theta )}{\sec \theta +\cos \theta }$ 
$N^r\&D^r$   dividing with  $\text{'}Tan\theta \text{'}\Rightarrow {\left(\frac{dy}{dx}\right)}^2=\frac{n^2{({\sec }^n\theta +{\cos }^n\theta )}^2}{{(\sec \theta +\cos \theta )}^2}$
 $=\frac{n^2(y^2+4)}{x^2+9}$