My conventions: a real-valued function on an open subset of $\mathbb{R}^n$ is smooth if its partial derivatives of all orders exist and are continuous. A real-valued function on an arbitrary subset $A\subseteq \mathbb{R}^n$ is smooth if it can be extended to a smooth function on an open subset containing $A$.
Let $f:\mathbb{R}^2\to\mathbb{R}$ be smooth. Furthermore, assume that $f$ is even with respect to its second input (i.e. $f(x,y)=f(x,-y)$. Is then the map \begin{align*}\mathbb{R}\times[0,\infty)&\to\mathbb{R}\\(x,y)&\mapsto f(x,\sqrt{y})\end{align*} smooth?
I've seen some similar questions here, but I haven't seen one treating the case with two variables.
No, this is false. As indicated in comment, this is essentially a one dimensional problem. Take $\varphi\in C^\infty$ be compactly supported, even and such that $\varphi = 1$ in a ball near $0$, and define $$ f(y) = \varphi(y)\,|y|^2 $$ Then $$ f(\sqrt y) = \varphi(y)\,|y| $$ is not differentiable at $x=0$ (I think your mistake was coming from the fact that you simplified $\sqrt{y^2} = y$ instead of $|y[$).