Note that

f(x) = 1 - 1/(x^2+2)

and

0 < 1/(x^2+2) <= 1/2.

Therefore, the image of f is [1/2, 1). If that's your co-domain, then yes, it's onto.

If the co-domain is anything larger than that, then no, it cannot be onto.

Note for x ∈R, 1/2 <= f(x) < 1. So, you decide !

If the function is defined by f : R -> [1/2,1) , in other words if its domain is real numbers and its codomain is real numbers between 1/2 and 1 , then the function is onto.

Needs more info. What is the co-domain?