To apply the inverse function theorem you need to prove that the derivative $Df(x)$ is invertible. This follows quite easily from the definition (with the limit $\lim_{h\to 0}\frac{f(x+h)-f(x)-Df(x)h}{\Vert h\Vert}=0$): If $y\neq 0$, then take $t$ very small, so that $\frac{\Vert f(x+ty)-f(x)-Df(x)ty\Vert}{\Vert ty\Vert}<\frac{1}{2}$. This implies
$$\frac{\Vert Df(x)y\Vert}{\Vert y\Vert}\geq\frac{\Vert f(x+ty)-f(x)\Vert}{\Vert ty\Vert}-\frac{\Vert f(x+ty)-f(x)-Df(x)ty\Vert}{\Vert ty\Vert}\geq1-\frac{1}{2}=\frac{1}{2},$$
so in particular $Df(x)y\neq 0$. Therefore $Df(x)$ is injective and, since it is a linear map, also surjective and hence invertible.
In particular, the Inverse Function Theorem states that $f$ is an open map with open image. The same proof as in https://math.stackexchange.com/a/3129225/58818 works in this case to prove that $f$ is surjective.