Here's what I have done so far:

$y=3x^2-(\frac{1}{x^2})$

$y'=\frac{6x^4+2}{x^3}$

So when x = 0, $y'$ wouldn't exist, right? And when $y'$ DNE at x, then x is a critical value?

Using 0 as a critical value, I'm getting that x < 0 as negative and x > 0 as positive, implying that there is a local min at x = 0, but the answer in the book doesn't support this.

EDIT: Looking over my notes again, I understand where my confusion is coming from, but I may still need some clarification. The picture you used (example a, a crazy looking function) has an asymptote where there is intuitively no max or min. But then soon after in the notes I have written down that x = c is a critical value when $f'(c)=0$ or DNE.