Is the boundary of a set defined using a smooth function also smooth?

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Let $f:B_\delta(y)\rightarrow \mathbb{R}^n$ be a smooth function for some $\delta>0$, $y\in\mathbb{R}^m$. Let us define the set $G:= f(B_\delta(y)) + \mathbb{R}^n_\geq$ (using the Minkowski sum). I want to investigate whether a point $\tilde{z}:=f(\tilde{x})$ for some $\tilde{x}\in\text{int}(B_\delta(y))$ (with int the interior of the set) is a smooth boundary point of $G$, i.e. whether there exists an open neighbourhood $U=U(z)$ and a smooth function $\rho:U\rightarrow \mathbb{R}$ that fulfils $$ U\cap G = \{z\in U: \rho(z)<0\} $$ and $$ \frac{d}{dz} \rho(z) \neq 0\ \forall z\in U. $$ Does this hold in general or do I need additional requirements? Could you point me to a book that is treating a similar problem? I am not very familiar with manifolds and therefore struggling with looking into the relevant literature.

Could you give me a hint on how to construct a function $\rho$?

Thank you for your help!