Answer
$+\sqrt { \frac{{ -5+\sqrt 33 }}{2}} $
$-\sqrt { \frac{{ -5+\sqrt 33 }}{2}} $
Work Step by Step
$y=\frac{x}{x^2+1}$
$ \frac{dy}{dt}=\frac {d (\frac{x}{x^2+1})} {dt}$
$ \frac{dy}{dt}=\frac {d (\frac{x}{x^2+1})} {dx} \frac{dx}{dt}$
$ \frac{dy}{dt}=\frac{(x^2+1) \frac{dx}{dx}-x\frac{d(x^2+1)} {dx}} {(x^2+1)^2} \frac{dx}{dt}$
$ \frac{dy}{dt}=\frac{(x^2+1) -2x^2} {(x^2+1)^2} \frac{dx}{dt}$
$ \frac{dy}{dt}=\frac {1 -x^2} {(x^2+1)^2} \frac{dx}{dt}$
............... eq (1)
Putting $\frac{dx}{dt} =3\times \frac{dy}{dt}$ in equation (1)
$ \frac{dy}{dt}=\frac {1 -x^2} {(x^2+1)^2} \times 3 \times \frac{dy}{dt}$
$ \frac {1}{3} =\frac {1 -x^2} {(x^2+1)^2} $
$ (x^2+1)^2=3 \times (1 -x^2)=3-3x^2 $
$x^4+2x^2+1+3x^2=3$
$x^4+5x^2-2=0$
Using quadratic formula
$x^2=\frac{-5+\sqrt 33}{2}$ Or $x^2=\frac{-5-\sqrt 33}{2}$
Or
$ x=\sqrt { \frac{-5+\sqrt 33}{2} } $ Or $ x=\sqrt { \frac{-5-\sqrt 33}{2} } $
