prove that $f:(0,1) \rightarrow \mathbb{R}$ such that $f(x)=\frac{2x-1}{x-x^2}$ is a bijection 
Let $f:(0,1)\rightarrow \mathbb{R}$ such that $f(x) = \frac{2x-1}{x-x^2}$. Prove that $f$ a bijection.

I proved that it's injective like this:
Let $x_1, x_2 \in (0,1)$ and $f(x_1)=f(x_2)$, then
$$\frac{2x_1-1}{x_1-x_1^2} = \frac{2x_2-1}{x_2-x_2^2} \implies (x_1-x_2)(2x_1x_2-x_1-x_2+1)=0$$
So $x_1=x_2$ or $(2x_1x_2-x_1-x_2+1)=0$.
The second case takes us to 
$x_2 = \frac{x_1-1}{2x_1-1}$, if $x_1>0.5$ then $x_2<0$ which is a contradiction. 
if $x_2<0.5$ then $1-x_1>1-2x_1$ and $x_2>1$ which is also a contradiction.
if $x_1=0.5$ then we find $0=0.5$ which is a contradiction, so the first case must be true. Which means $f$ is injective.
But how can I prove that $f$ is surjective? (I can't find an inverse function for $f$).
 A: let us prove $f$ is surjective.
take $y\in\mathbb R$
we seek $x\in(0,1)$ such that
$2x-1=y(x-x^2)$ 
which can be written as follows
$yx^2+(2-y)x-1=0$
if $y=0$ then $x=\frac{1}{2}$
assume now $y\neq0$.
the discriminant is
$\delta=4+y^2-4y+4y=y^2+4>0$
thus $x$ exists and it is given by
$x=\frac{y-2 +\sqrt{\delta}  }{ 2y }$
or
$x=\frac{y-2-\sqrt{\delta}}{2y}$
Qed.
A: $$f(x)=\frac{2x-1}{x-x^2}=\frac{x+(x-1)}{x(1-x)}=\\\frac{x-(1-x)}{x(1-x)}=\\\frac{1}{1-x}-\frac{1}{x}\\$$
$$x \in(0,1)  \to g(x)=\frac{1}{1-x} \space \to g'=\frac{1}{(1-x)^2}>0$$
$$x \in(0,1)  \to h(x)=\frac{1}{-x} \space \to h'=\frac{1}{(x)^2}>0$$
$$x \in(0,1)  \to g(x)+hx)=f(x) \space \to f'>0$$ so $f(x)$is one to one function $x \in(0,1)$ $\color{red} {*}$
$$x \to 0^+ \to f(x)\to-\infty\\x \to 1^- \to f(x)\to+\infty\\ $$
f(x) is continuous
 $\in (0,1)$ 
so $f(x)$ is surjective function in $(0,1)\\$ $\color{red} {**}$  $$\color{red} {**},\color{red} {*}$$
 f(x) is bijective  in$(0,1)$
A: here is a better method.
the numerator of the derivative of $f$ is
$2(x-x^2)-(2x-1)(1-2x)=$
$2x^2-2x+1=$
$x^2+(x-1)^2>0$.
so $f$ is continuous and strictly increasing at $(0,1)$,  thus it is a bijection from $(0,1)$ to 
$(\lim_{x\to0^+}f(x),\lim_{x\to1^-}f(x))=$
$(-\infty,+\infty)$.
