We have: $x < 0$

Also, $x > 1/x$

Thus, x must be a negative fraction between 0 and -1

For example: $x = -1/2 => 1/x = -2 => -1/2 > -2$

Thus: $-1 < x < 0$

Working with the options:

A. Since $x$ is a negative fraction between 0 and -1, $x^2$ must be a positive fraction between 0 and 1 => $1<x^2$ is False

B. Since $x$ is negative, and $x^2$ is positive, $x^2$ must be greater than x => $x^2<x$ is False

C. Since $x$ is a fraction between 0 and -1, $x^3$ will be a negative fraction also lying between 0 and -1 => $-1<x^3<0$ is True

D. Since $x$ lies between 0 and -1, $1/x$ must be less than -1 => $\frac{1}{x} >-1$ is False

E. Since $x$ is a fraction between 0 and -1, $x^3$ will be a negative fraction but having a smaller magnitude, i.e. $x^3$ is greater than $x$ => $x^3<x$ is False


Answer C