$\mathrm{If} \mathrm{x} = \frac{2\mathrm{at}}{1+{\mathrm{t}}^2}, \mathrm{y}=\frac{\mathrm{b}\left(1-{\mathrm{t}}^2\right)}{1+{\mathrm{t}}^2},$ then $\frac{\mathrm{dy}}{\mathrm{dx}}$

Given $\mathrm{x} = \frac{2\mathrm{at}}{1+{\mathrm{t}}^2}, \mathrm{y}=\frac{\mathrm{b}\left(1-{\mathrm{t}}^2\right)}{1+{\mathrm{t}}^2},$

Differentiate both $x$ and $y$ with respect to $t$

$\begin{array}{rcl}\frac{dx}{dt}& =& \frac{d}{dt}\left(\frac{2at}{1+t^2}\right)\\ & =& 2a\frac{\left(1+t^2\right)1-t\left(2t\right)}{{\left(1+t^2\right)}^2}\\ & =& 2a\frac{1+t^2-2t^2}{{\left(1+t^2\right)}^2}\\ & =& 2a\frac{1-t^2}{{\left(1+t^2\right)}^2}\end{array}$

$\begin{array}{rcl}\frac{dy}{dt}& =& \frac{d}{dt}\left( \frac{\mathrm{b}\left(1-{\mathrm{t}}^2\right)}{1+{\mathrm{t}}^2}\right)\\ & =& b\frac{\left(1+t^2\right)\left(-2t\right)-\left(1-t^2\right)2t}{{\left(1+t^2\right)}^2}\\ & =& \frac{-2bt\left(1+t^2+1-t^2\right)}{{\left(1+t^2\right)}^2}\\ & =& \frac{-4bt}{{\left(1+t^2\right)}^2}\end{array}$

Use the concept of parametric differentiation 

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{dy}{dt}·\frac{dt}{dx}\\ & =& -\frac{4bt}{{\left(1+t^2\right)}^2}·\frac{{\left(1+t^2\right)}^2}{2a\left(1-t^2\right)}\\ & =& \frac{-2bt}{a\left(1-t^2\right)}\\ & =& \frac{-b{\displaystyle \frac{x}{a}}}{a{\displaystyle \frac{y}{b}}}\\ & =& \boxed{\frac{-b^{2}x}{a^{2}y}}\end{array}$