By definition, the limit of a function at a point is defined as
$$\exists L:\left( {\forall \varepsilon > 0,\exists \delta > 0:\left( {\forall x,0 < \left| {x - a} \right| < \delta \to \left| {f(x) - L} \right| < \varepsilon } \right)} \right)$$
and its negation will be (see this post)
$$\forall L,\exists \varepsilon > 0:\left( \forall \delta > 0,\exists x:(0 < \left| x - a \right| < \delta \wedge |f(x) - L| \ge \varepsilon \right))$$
So for your special case we shall prove that
$$\forall L,\exists \varepsilon > 0: ( \forall \delta > 0,\exists x:(0 < \left| x \right| < \delta \wedge |\frac{1}{x^2} - L| \ge \varepsilon )$$
You can see this as a game! Your opponent chooses $L$ then you select an $\epsilon$. Next, the opponent chooses $\delta$ and then you choose the $x$. You should be wise enough to make proper choices so that you will win the game no matter what your opponent chooses. So your choices must satisfy the conditions $\left| x \right| < \delta \wedge |\frac{1}{x^2} - L| \ge \varepsilon$ regardless of opponent's choices. So, let us play!
The opponent makes a choice for $L$. we choose $\varepsilon$ to be $1$. Then the opponent selects a $\delta$ and we should make our final choice for $x$ such that
$$\left| x \right| < \delta \wedge |\frac{1}{x^2} - L| \ge 1$$
we can choose $x_n=\frac{1}{\sqrt{n+L}}$ and our conditions turns to be
$$\left| \frac{1}{\sqrt{n+L}} \right| < \delta \wedge |n| \ge 1$$
So, no matter what the opponent chooses, we will select large enough values for $n$ such that the above conditions are satisfied and we win the game! Hence, the limit does not exist!