Answer
The asymptotes are:
1) Line $x=0$ is vertical asymptote from the left and $y$ goes to $+\infty$;
2) Line $x=0$ is vertical asymptote from the right and $y$ goes to $+\infty$;
3) Line $y=0$ is horizontal asymptote when $x\to-\infty$;
4) Line $y=0$ is horizontal asymptote when $x\to +\infty$.
Work Step by Step
From the graph below we suspect that the asymptotes are:
1) Line $x=0$ is vertical asymptote from the left and $y$ goes to $+\infty$;
2) Line $x=0$ is vertical asymptote from the right and $y$ goes to $+\infty$;
3) Line $y=0$ is horizontal asymptote when $x\to-\infty$;
4) Line $y=0$ is horizontal asymptote when $x\to +\infty$.
Let's prove this:
1) $$\lim_{x\to0^-}\frac{\cos^2x}{x^2}=\left[\frac{\cos^2 0^-}{(0^-)^2}\right]=\left[\frac{1}{0^+}\right]=+\infty$$
which is what is needed.
2) $$\lim_{x\to 0^+}\frac{\cos^2 x}{x^2}=\left[\frac{\cos^20^+}{(0^+)^2}\right]=\left[\frac{1}{0^+}\right]=+\infty$$
which is what is needed.
3) $$\lim_{x\to -\infty}\frac{\cos^2x}{x^2}=\left[\frac{\text{something that is always between $-1$ and $1$}}{(-\infty)^2}\right]=\left[\frac{\text{something that is always between $-1$ and $1$}}{+\infty}\right]=0.$$
which is what is needed.
4) $$\lim_{x\to +\infty}\frac{\cos^2x}{x^2}=\left[\frac{\text{something that is always between $-1$ and $1$}}{(+\infty)^2}\right]=\left[\frac{\text{something that is always between $-1$ and $1$}}{+\infty}\right]=0.$$
which is what is needed.
