Đáp án:

\[\left[ \begin{array}{l}
x > 1\\
 - 1 < x < 0
\end{array} \right.\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
\left( {x \ne 0} \right)\\
3\left( {{x^2} - \frac{1}{{{x^2}}}} \right) < 2\left( {{x^3} - \frac{1}{{{x^3}}}} \right)\\
 \Leftrightarrow 3\left( {x - \frac{1}{x}} \right)\left( {x + \frac{1}{x}} \right) < 2\left( {x - \frac{1}{x}} \right)\left( {{x^2} + x.\frac{1}{x} + \frac{1}{{{x^2}}}} \right)\\
 \Leftrightarrow 3\left( {x - \frac{1}{x}} \right)\left( {x + \frac{1}{x}} \right) < 2\left( {x - \frac{1}{x}} \right)\left[ {\left( {{x^2} + 2.x.\frac{1}{x} + \frac{1}{{{x^2}}}} \right) - 1} \right]\\
 \Leftrightarrow 3\left( {x - \frac{1}{x}} \right)\left( {x + \frac{1}{x}} \right) < 2\left( {x - \frac{1}{x}} \right)\left[ {{{\left( {x + \frac{1}{x}} \right)}^2} - 1} \right]\\
 \Leftrightarrow \left( {x - \frac{1}{x}} \right)\left[ {2{{\left( {x + \frac{1}{x}} \right)}^2} - 2 - 3\left( {x + \frac{1}{x}} \right)} \right] > 0\\
 \Leftrightarrow \left( {x - \frac{1}{x}} \right)\left( {x + \frac{1}{x} - 2} \right)\left( {x + \frac{1}{x} + \frac{1}{2}} \right) > 0\\
TH1:\,\,\,\left\{ \begin{array}{l}
x - \frac{1}{x} > 0\\
\left( {x + \frac{1}{x} - 2} \right)\left( {x + \frac{1}{x} + \frac{1}{2}} \right) > 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\frac{{{x^2} - 1}}{x} > 0\\
\left[ \begin{array}{l}
x + \frac{1}{x} > 2\\
x + \frac{1}{x} <  - \frac{1}{2}
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x > 1\\
 - 1 < x < 0
\end{array} \right.\\
\left[ \begin{array}{l}
\frac{{{x^2} - 2x + 1}}{x} > 0\\
\frac{{{x^2} + \frac{1}{2}x + 1}}{x} < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x > 1\\
 - 1 < x < 0
\end{array} \right.\\
\left[ \begin{array}{l}
x > 0\\
x < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x > 1\\
 - 1 < x < 0
\end{array} \right.\\
TH2:\,\,\,\left\{ \begin{array}{l}
x - \frac{1}{x} < 0\\
\left( {x + \frac{1}{x} - 2} \right)\left( {x + \frac{1}{x} + \frac{1}{2}} \right) < 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\frac{{{x^2} - 1}}{x} < 0\\
x + \frac{1}{x} >  - \frac{1}{2}\\
x + \frac{1}{x} < 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x <  - 1\\
0 < x < 1
\end{array} \right.\\
\frac{{{x^2} + \frac{1}{2}x + 1}}{x} > 0\\
\frac{{{x^2} - 2x + 1}}{x} < 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x <  - 1\\
0 < x < 1
\end{array} \right.\\
x > 0\\
x < 0
\end{array} \right.\,\,\left( {vn} \right)
\end{array}\)

Vậy \(\left[ \begin{array}{l}
x > 1\\
 - 1 < x < 0
\end{array} \right.\)