let be a function so
$ f(x)= x^{2}+h(x) $ , here $h(x) $ is a function so $ h(x) = O(logx)$
it is clear that $ f(x) \sim x^{2} $ for big 'x' so the inverse
$ f^{-1}(x)\sim x^{1/2}$ for big 'x' is this correct, does this mean that the function $ f(x)$ has an inverse and that this inverse will be asymptotic to $ x^{1/2} $ ?
No, for instance, if $$ h(x)=\begin{cases}x & \text{if $\lvert x\rvert<2$}\\ 0&\text{else}\end{cases} $$ then $f(0)=f(-1)$, hence $f$ is not invertible. This counterexample could be turned into a continuous or smooth one easily.