Your idea would work fine. If you can prove that f(x) is always increasing or decreasing, you have proven that f(x) has an inverse.
$\displaystyle f(x)$ increases when $\displaystyle f'(x)>0$
$\displaystyle f(x)$ decreases when$\displaystyle f'(x)<0$
$\displaystyle f'(x) = 3x^2 -1$
Now we need to check if $\displaystyle f'(x)$ changes sign (crosses the x-axis). If it does change sign, the $\displaystyle f(x)$ does
not have an inverse. First, we check for zeros of the derivative:
$\displaystyle 3x^2 - 1 = 0$
$\displaystyle x^2 = \frac{1}{3}$
Zeros of $\displaystyle f'(x)$ are:
$\displaystyle x=\frac{1}{\sqrt3}, -\frac{1}{\sqrt3}$
This does not conclusively prove anything though, because it only says that the derivative has zeros, but not that it changes sign. To check if it changes sign, just figure out if the derivative is positive or negative in the regions between the zeros:
$\displaystyle f'(0)=-1$ [substitute 0 into the derivative equation] Since the derivative is negative, $\displaystyle f(x)$ is decreasing at x=0.
$\displaystyle f'(1)=2$ Since the derivative is positive, $\displaystyle f(x)$ is increasing at x=1.
This proves that $\displaystyle f(x)$ does not have an inverse because it is increasing at some point, but decreasing at another.
Hope I helped