Suppose $f\colon \mathbb R ^2 \to \mathbb R$ is defined by $(x_1, x_2) \mapsto x_1+x_2 + x_1x_2$. Let $E \subset \mathbb R$ be a connected set containing $0$ and $E$ is not the singleton $0$. Is $f^{-1}(E)$ connected?
I asked a similar question here. I tried to follow one of the answers to construct a path to $0$ by $t \mapsto t(x_1, x_2),$ but this does not seem to work since there is a sum of different order of $t$.
Hint. Consider the connected set $E=[-1/2,+\infty)$ which contains $0$ in its interior and it is not a singleton. Then $S:=f^{-1}([-1/2,+\infty))$ is the set of points $(x_1,x_2)\in \mathbb R ^2$ such that $$x_1+x_2 +x_1x_2\geq -1/2.$$ Is $S$ connected? Note that $P:=(0,0)\in S$ and $(-2,-2)\in S$. Let $\gamma$ be a continuous path from $P$ to $Q$ then, by continuity, $\gamma$ will intersect the line $x_1=-1$ at some point $R$. Does $R$ belong to $S$?