Answer
$$y'=-\frac{y}{1+x^2}$$
Work Step by Step
$$ye^{\tan^{-1}x}=2$$
Using implicit differentiation, we differentiate both sides of the equation with respect to $x$: $$y'e^{\tan^{-1}x}+y(e^{\tan^{-1}x})'=0$$
$$y'e^{\tan^{-1}x}+y(e^{\tan^{-1}x})(\tan^{-1}x)'=0$$
$$y'e^{\tan^{-1}x}+y(e^{\tan^{-1}x})\times\frac{1}{1+x^2}=0$$
$$y'e^{\tan^{-1}x}+\frac{ye^{\tan^{-1}x}}{1+x^2}=0$$
Next, we need to separate elements with $y'$ and those without $y'$ into 2 sides of the equation:
$$y'e^{\tan^{-1}x}=-\frac{ye^{\tan^{-1}x}}{1+x^2}$$
Finally, calculate for $y'$: $$y'=-\frac{ye^{\tan^{-1}x}}{(1+x^2)e^{\tan^{-1}x}}$$
$$y'=-\frac{y}{1+x^2}$$
