(a) \[\frac{dy}{dx}=\frac{-2\sin\theta\cos\theta}{\sec^2\theta}=-2\sin\theta\cos^3\theta.\]

(b) Eliminating the parameter gives \(y=1/(1+x^2)\), whose gradient is \(G(x)=-2x/(1+x^2)^2\). Then \[G^{\prime}(x)=\frac{6x^2-2}{(1+x^2)^3}.\] Its critical values are \(x=\pm1/\sqrt3\). The positive maximum of the gradient is on the left branch, so \(\boxed{x_P=-1/\sqrt3}\).