Calculation of $\max$ and $\min$ value of $f(x) = \frac{x(x^2-1)}{x^4-x^2+1}.$ 
Calculation of $\max$ and $\min$ value of $$f(x) = \frac{x(x^2-1)}{x^4-x^2+1}$$

My try: We can write $$f(x) = \frac{\left(x-\frac{1}{x}\right)}{\left(x^2+\frac{1}{x^2}\right)-1} = \frac{\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+1}$$ 
Now put $\left(x-\frac{1}{x}\right)=t,x\ne0$. Then we get $$f(t) = \frac{t}{t^2+1} = \frac{1}{2}\left(\frac{2t}{1+t^2}\right)$$
Now put $t=\tan \theta$. Then $$f(\theta) = \frac{1}{2}\frac{2\tan \theta}{1+\tan^2 \theta} = \frac{1}{2}\sin 2\theta$$
So we get $$-\frac{1}{2}\leq f(\theta)\leq \frac{1}{2}\Rightarrow f(\theta)\in \left[-\frac{1}{2}\;,\frac{1}{2}\right]$$
My question is: Is my solution right? If not, then how can we solve it?

The denominator 'factorises', giving us this:
$$f(x) = {x(x^2-1) \over (x^2-1)^2 + x^2}
$$
We can establish this is less than $1/2$ fairly easily:
$${uv \over {u^2+v^2}} \le {1 \over 2}
\iff u^2 + v^2 - 2uv  \ge 0
\iff (u-v)^2 \ge 0
$$
($u,v$ not both 0, but this holds for us.)
And equality is obtained when $u=v$, i.e. $x=x^2-1$, and this has a positive solution (which is in fact the golden ratio).
I think a sketch is instructive. Note that: 


*

*The denominator never vanishes (consider it as a quadratic in $x^2$) 

*as $x \rightarrow \infty$, $f(x) \rightarrow 0$

*$f(x)$ is an odd function


These along with the extremal value work we've just done should be enough for a sketch.
