$\text{ If }x=\frac{1-t^2}{1+t^2}\text{ and }y=\frac{2t}{1+t^2}$ then $\frac{dy}{dx}$ is

let $\mathrm{x}=\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}\text{ and }\mathrm{y}=\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}$
put $\mathrm{t}=\tan \theta$ we get 
$x=\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\text{ and }y=\frac{2\tan \theta }{1+{\tan }^2\theta }$
$x=\cos 2\theta \text{ and }y=\sin 2\theta$
$\begin{array}{r}\frac{\mathrm{dx}}{\mathrm{d}\theta }=-2\sin 2\theta \text{ and }\frac{\mathrm{dy}}{\mathrm{d}\theta }=2\cos 2\theta \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{d}\theta }\times \frac{\mathrm{d}\theta }{\mathrm{dx}}=-\frac{\cos 2\theta }{\sin 2\theta }=-\frac{\mathrm{x}}{\mathrm{y}}\end{array}$