= Solution
The <implicit function theorem> states that if $F(x_0,y_0)=0$ and the partial derivative $D_yF(x_0,y_0)$ is an <invertible linear map>, then near $(x_0,y_0)$ the zero set of $F$ is uniquely the graph $y=g(x)$ of a continuously differentiable function.
If $f$ is a <differentiable map>[differentiable] bijection with differentiable inverse $g$, the <chain rule> applied to $g\circ f$ and $f\circ g$ gives
$$
Dg_{f(x)}Df_x=I,
\qquad
Df_xDg_{f(x)}=I.
$$
Thus $Df_x$ is an isomorphism with inverse $Dg_{f(x)}$.
If a continuously differentiable map $F:\mathbb R^n\to\mathbb R^n$ has invertible derivative everywhere, the <inverse function theorem> makes it a local diffeomorphism. In particular it is an <open map>, so its image is open. Its image need not be closed: $x\mapsto\arctan x$ has nonzero derivative everywhere and image $(-\pi/2,\pi/2)$.
For the given map of <elementary symmetric polynomial>[elementary symmetric polynomials],
$$
DF=
\begin{pmatrix}
1&1&1\\
y+z&x+z&x+y\\
yz&xz&xy
\end{pmatrix},
$$
and direct evaluation of the <determinant> gives
$$
\det DF=(x-y)(y-z)(z-x).
$$
Hence the <critical set> is
$$
\boxed{C=\{x=y\}\cup\{y=z\}\cup\{z=x\}}.
$$
Its complement is the <Zariski-open set> on which the three coordinates are pairwise distinct. Each point has one of the six possible strict coordinate orderings, and each ordering defines a nonempty convex open region. A <continuous path> cannot change an ordering without crossing $C$. Therefore $\mathbb R^3\setminus C$ has exactly
$$
\boxed{3!=6}
$$
<connected component>[connected components].
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