Solving $\frac{1}{x} + \frac{1}{1-x} > 0$. Where's my error? 
For which $x$ is the following true?
$$\frac{1}{x} + \frac{1}{1-x} > 0$$

I’ve worked through the question a couple ways and got the correct answer, but the initial approach has a flaw and I’m not sure where in which step it is.
I got the correct answer by adding the two numbers to get $\frac{1}{x-x^2} > 0$ and going from there.
However my first attempt went like:
$$\begin{align}\frac{1}{x} &> - \frac{1}{1-x} \tag1\\[4pt]
\Rightarrow \qquad \frac{1-x}{x} &> -1 \tag2\\[4pt]
\Rightarrow \qquad 1-x &> -x \tag3 \\[4pt]
\Rightarrow \qquad x-1 &< x \tag4 
\end{align}$$
For which the answer is all $x$. However, the correct answer, which is clear from the first approach is $0 < x < 1$.
I’m new to working with inequalities and I figure there is just a simple property of them I’m being ignorant about which means one of my rearrangements isn’t true.
All help appreciated. Thanks.
 A: "I figure there is just a simple property of them I’m being ignorant about which means one of my rearrangements isn’t true": that's it.
Let us take a few examples :

*

*$\color{blue}5<\color{green}8$ and $\color{blue}5\times2<\color{green}8\times2$

*$\color{blue}{-5}<\color{green}{\frac83}$ and $\color{blue}{-5}\times2<\color{green}{\frac83}\times2$

*$\color{blue}{-8}<\color{green}{-5.3}$ and $\color{blue}{-8}\times2<\color{green}{-5.3}\times2$
But

*

*$\color{blue}5<\color{green}8$ and $\color{blue}5\times{(-2)}\color{red}>\color{green}8\times(-2)$

*$\color{blue}{-5}<\color{green}{8}$ and $\color{blue}{-5}\times(-2)\color{red}>\color{green}{8}\times(-2)$

*$\color{blue}{-8}<\color{green}{-5}$ and $\color{blue}{-8}\times(-2)\color{red}>\color{green}{-5}\times(-2)$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{{\displaystyle #1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
& \color{#44f}{{1 \over x} + {1 \over 1 -x} > 0}
\sr{x\ \not\in\ \braces{0,1}}\implies
x^{2}\pars{1 - x}^{2}\
\overbrace{{1 \over x\pars{1 - x}}}
^{\ds{= 1/x + 1/\pars{1 - x}}} > 0
\\[5mm] & \implies x\pars{1 - x} >0 \implies
x\pars{x - 1} < 0 \implies
\bbx{\color{#44f}{0 < x < 1}}
\end{align}
A: you must check the inequality on the three intervals on which the expression is defined. These are $(-\infty,0) , \ (0,1), \ $and   $(1,\infty)$. of you take one test  point from each interval you will find that the inequality holds only when $0<x<1$
The flaw in your initial approach  is when you multiply by  $x-1$. If $x-1<0$ the direction of the inequality changes
also see Solving $\frac1x + \frac1{1-x} > 0$ two ways gives different results, one of which is simply $0<1$. What does this mean?
A: When you multiply by $1 - x$, if $1 - x < 0$ you flip the direction of the inequality.
Then inequality $(2)$ would be backwards;
it should be $\frac{1-x}{x} < -1$ in that case.
Again, when you multiply by $x$, if $x < 0$ you flip the direction of the inequality.
In no case can either $x$ or $1 - x$ be zero, because then the original inequality would be comparing something with an undefined result.
If $x$ and $1 - x$ are both positive, all your steps are correct and moreover you can make the implications bidirectional, so the true statement at the end implies your initial statement was true; that is, it is sufficient for $x$ and $1 - x$ both to be positive.
If $x$ and $1 - x$ are both negative, your first two steps are wrong; to make them correct you need to reverse the $>$ to $<$ in inequality $(2).$ But with that correction, you can make the implications bidirectional, so the true statement at the end implies your initial statement was true; that is, it is sufficient for $x$ and $1 - x$ both to be negative.
If one of $x$ and $1 - x$ is positive and the other is negative, the direction of the inequality gets flipped in only one step, so in the final step (correcting the directions of the inequalities) you have $x - 1 > x.$
Since this is always false, you can never satisfy the original inequality
when $x$ and $1 - x$ have opposite signs.
So the original inequality is true exactly when $x$ and $1 - x$ have the same sign
(both positive or both negative), that is, when
$$ x (1 - x) > 0. $$
