Range of a function, with contradictory restriction

You have the equation $y = \frac{x}{x^2 + 1}$, which you then simplified to $x^2y + y - x = 0$. Then, solving this for $x$ gives you $\frac{1 \pm \sqrt{1-4y^2}}{2}$, which is all good, very good, until...

If $y = 0$, then the above is not even a quadratic equation, so you cannot approach it through the quadratic formula! Instead, you must deal with such a case separately, by setting $y = 0$, which gives $x = 0$.

Moral of the (real) story : The quadratic formula applies only when the quadratic coefficient is non-zero. Otherwise, such a quadratic simplifies to a linear (possibly, even better) equation which can be solved.

Note that $f(0) = 0$ and $f(x)\to0$ as $x\to\infty$ and $f(x)>0$ when $x>0,$ and $f$ is everywhere continuous. That implies that as $x$ goes from $0$ to $\infty,$ then $f(x)$ must go up from $0$ and ultimately back down to $0.$ Thus $f$ has an absolute maximum somewhere to the right of $0.$ And $f(x)$ must assume all intermediate values between $0$ and that maximum.

Next: $$ \frac x {x^2+1} = \overbrace{ \frac 1 {x + \frac 1 x} = \frac 1 {\left( x - 2 + \frac 1 x \right) + 2} }^\text{This is a sort of completion of the square} = \frac 1 {\left( \sqrt x - \frac 1 {\sqrt x} \right)^2 + 2}. $$ This is equal to $1/2$ when the square is $0;$ otherwise it is less than $1/2.$

Thus $f$ maps $[0,\infty)$ to $[0,1/2].$

Since $f$ is an odd function, it therefore maps $(-\infty,0]$ to $[-1/2,0].$