Is the following true? If so, I would be grateful for a reference that contains such a result and its proof. 


Let $f:\mathbb{R}^d\rightarrow \mathbb{R}$ be a real analytic function, and $V:=\{\mathbf{x}\in \mathbb{R}^d: f(\mathbf{x}) =0\}$ be its variety. Then there exists a decomposition $V=V\_{d-1}\cup \dots \cup V\_0$, where some of the $V\_k$s may be empty, and where the $V\_k$ are real analytic submanifolds of $\mathbb{R}^d$ of dimension $k$, admitting real analytic parametrizations as follows: For every $\mathbf{x}\_0\in V\_k$, there exists a neighbourhood $U$ of $\mathbf{x}\_0$ in $\mathbb{R}^d$ and a neighbourhood $U'$ of $\mathbf{0}\in \mathbb{R}^d$, with a homeomorphism $\varphi:U'\rightarrow U$ which is real analytic, and sends $\mathbb{R}^k\times \mathbb{R}^{d-k}\supset U' \owns (\mathbf{z},\mathbf{w})\mapsto \varphi(\mathbf{z},\mathbf{w})\in U$, such that 
$V\_k\cap U=\{\varphi(\mathbf{z},\mathbf{0}): \mathbf{z}\in \mathbb{R}^k \text{ such that }(\mathbf{z},\mathbf{0})\in U'\}$.



Decomposition of a real analytic variety
You may want to look at Lojasiewicz's structure theorem; for a statement and proof (I won't reproduce it here since the complete theorem statement is over a page long) see Chapter 6, section 3 of [Krantz's *A Primer of Real Analytic Functions*](https://www.springer.com/gp/book/9780817642648). 


(The theorem as stated there is not exactly how you phrased it above, but I think it is stronger and some minor massaging can get what you want.)


