Related Rates. Finding all values of x. A particle is moving along the curve $y=\frac{x}{x^2+1}$. Find all values of x in which the rate of change of x wrt time is 3 times that of y.
So far, I have the derivative function, which I believe is: 
$$\frac{dy}{dt}=\frac{dx}{dt}\frac{1-x^2}{(x^2+1)^2}$$
I'm not sure if this is correct, and I'm not sure where to go from here. I assume I need to solve for x by substituting $\frac{dx}{dt}=3$ and $\frac{dy}{dt}=1$, but when I do I get:
$$x^4+5x^2-2=0$$
I can't seem to find a factor for this. Am I on the right track? Any help would be appreciated 
 A: Your initial setup is correct: $\dfrac {dy}{dt}$ in terms of $\dfrac {dx}{dt}$ and $x$.
What you also need to use is the relationship: $$\frac{dx}{dt} = 3\left(\frac{dy}{dt}\right)\iff \dfrac {dy}{dt} = \frac 13 \frac{dx}{dt}$$
Now then, solve $$\frac{dy}{dt}=\frac{dx}{dt}\frac{1-x^2}{(x^2+1)^2} \iff \frac 13\left(\frac {dx}{dt}\right)=\frac{dx}{dt}\frac{1-x^2}{(x^2+1)^2}$$ for the value $x$. You'll see that the factor $\dfrac{dx}{dt}$ assuming $\dfrac{dx}{dt} \neq 0$, leaving you with your degree four polynomial in $x$.
And indeed, you've found that formula. There here will be $2$ solutions in $x^2$, one of which needs to be thrown out, since $x^2$ can not be negative. (You can simply use the quadratic formula to solve for $x^2$. For the valid value of $x^2$, there will be two solutions.
A: The rate of change of x wrt time is 3 times that of y:
$${{dx} \over {dt}} = 3{{dy} \over {dt}}$$
Consider the differentiation of $y=\frac{x}{x^2+1}$ :
$$\frac{dy}{dx}=\frac{1-x^2}{(x^2+1)^2}$$
So if I divide both sides by $dt$, we get:
$$\frac{dy}{dt}=\frac{dx}{dt}\frac{1-x^2}{(x^2+1)^2}=\frac{3(1-x^2)}{(x^2+1)^2}\frac{dy}{dt}$$
Eliminate $\frac{dy}{dt}$:
$$
x^4+5x^2-2=0 \Longrightarrow x^2=\frac{\sqrt{33}-5}{2}\Longrightarrow x =  \pm \sqrt {{{\sqrt {33}  - 5} \over 2}} 
$$
