Draw the graph of `y=x^(2)+(1)/(x^(2)), x !=0`.

Explanation

We have `f(x)=x^(2)+(1)/(x^(2)), x !=0`.
Clearly , x=0 is the vertical asymptote.
Curve has no horizontal and oblique asymptotes.
Since the function is even, draw the graph of `y=f(x)` for `x gt 0` first.
`underset(x rarr 0^(+))lim (x^(2)=(1)/(x^(2)))=oo` and `underset(x rarr oo)lim(x^(2)+(1)/(x^(2)))=oo`
`f'(x)=2x-(2)/(x^(3)),f''(x)=2+(6)/(x^(4))`
`f'(x)=0 rArr x^(4)=1 rArr x=1`, which is point of minima as `f''(1)=8 gt 0`
`f(1)=2`
So the graph of `y=f(x)` for `x gt 0` is as shown in the following figure.

Since `y=f(x)` is an even function , the graph for `R-{0}` is as shown in the following figure .