Show that $y = \frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Prove, using an algebraic method,that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Hence, determine the minimum and maximum points $y=\frac{2x}{x^2 +1}$ . 
What I tried:
Firstly, I thought of using partial fractions but since $x^2 +1 =(x-i)(x+i)$, I don't think it is possible to show using partial fractions.
Secondly, decided to use differentiation
$y=\frac{2x}{x^2 +1}$
$\frac {dy}{dx} = \frac {-2(x+1)(x-1)}{(x^2 +1)^2 }$
For stationary points:
$\frac {dy}{dx} = 0$
$\frac {-2(x+1)(x-1)}{(x^2 +1)^2 } = 0$
$x=-1$ or $x=1$
When $x=-1,y=-1$
When $x=1,y=1$
Therefore, this implies that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.
^I wonder if this is the correct method or did I leave out something?
The third way was using discriminant
Assume that $y=\frac{2x}{x^2 +1}$ intersects with $y=-1$ and $y=1$
For $\frac{2x}{x^2 +1} = 1$,
$x^2 -2x+1 = 0$
Discriminant = $ (-2)^2  -4(1)(1) = 0 $
For $\frac{2x}{x^2 +1} = -1$,
$x^2 +2x+1 = 0$
Discriminant = $ (2)^2  -4(1)(1) = 0 $
So, since $y=\frac{2x}{x^2 +1}$ touches $y=-1$ and $y=1$, $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.
Is the methods listed correct?Is there any other ways to do it?

Your second and third approaches are workable, but they lack the part proving that the fraction otherwise is between $-1$ and $1$.
For the second you use the sign of the derivate between and beyond the stationary points, especially beyond. You then can see that for $x>1$ the function is decreasing and for $x<-1$ too (and for $-1\le x\le 1$ it's increasing.
For the discriminant you consider for a given candidate value of $y$ the equation $2x/(x^2+1) = y$ which becomes
$$yx^2 - 2x + y = 0$$
which has the discriminant $(-2)^2 - 4y^2 = 4(1-y^2)$ and the equation only have solutions if $1-y^2=(1-y)(1+y)\ge 0$ which means that the factors are both positive or both negative (but they can't be both negative). So $1-y \ge 0$ and $1+y\ge 0$ that is $-1\le y\le 1$.
