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This non-convexity idea can be generalized, with $$P(x,y)=P_{k,h}(x,y) :=\prod_{j=0}^{2k-1}\Big(x\cos\frac{\pi j}k-y\sin\frac{\pi j}k\Big)^2+h(x^{4k}+y^{4k})$$ for natural $k$ and real $h>0$. Here is the picture of the curve $P_{k,h}^{-1}(\{1\})$ for $k=5$ and $h=(3/10)^{4k}$:

One may also note that in general the level sets $P^{−1}([0,c])$ will not be convex, even if $P$ is a positive elliptic homogeneous polynomial. E.g., take $P(x,y)=(x−y)^2(x+y)^2+h(x^4+y^4)$ for a small enough $h>0$. Here is the picture of this level set for $c=1$ and $h=1/10$:

Clearly, the shape of this level set does not depend on $c>0$.

For any $d_*\in(0,1)$ and any real $D>0$, let $\mathcal P_{n,d_*,D}$ denote the set of all polynomials $p(x)=\sum_{j=0}^n d_j x^j$ such that $d_n\ge d_*$ and $\sum_{j=0}^n|d_j|\le D$. Then it is not hard to see that there is a real $c_*(n,d_*,D)>0$, depending only on $n,d_*,D$, such that for any polynomial $p(x)=\sum_{j=0}^n d_j x^j$ in $\mathcal P_{n,d_*,D}$ and for all real $c\ge c_*(n,d_*,D)$ the equation $p(x)=c$ has exactly two roots $x_\pm:=x_\pm(c)$ such that $x_-<0<x_+$ and, moreover, \begin{equation*} x_\pm=\pm\Big(\frac c{d_n}\Big)^{1/n}-(1+o(1))\frac{d_{n-1}}{nd_n} \tag{1} \end{equation*} uniformly over all polynomials $p(x)=\sum_{j=0}^n d_j x^j$ in $\mathcal P_{n,d_*,D}$ (as $c\to\infty$). This uniformity can be obtained by refining this reasoning.
Moreover, without loss of generality, for all $p\in\mathcal P_{n,d_*,D}$ and real $c\ge c_*(n,d_*,D)$ we have $p'(x_\pm)\ne0$.

For any $d_*\in(0,1)$ and any real $D>0$, let $\mathcal P_{n,d_*,D}$ denote the set of all polynomials $p(x)=\sum_{j=0}^n d_j x^j$ such that $d_n\ge d_*$ and $\sum_{j=0}^n|d_j|\le D$. Then it is not hard to see that there is a real $c_*(n,d_*,D)>0$, depending only on $n,d_*,D$, such that for any polynomial $p(x)=\sum_{j=0}^n d_j x^j$ in $\mathcal P_{n,d_*,D}$ and for all real $c\ge c_*(n,d_*,D)$ the equation $p(x)=c$ has exactly two roots $x_\pm:=x_\pm(c)$ such that $x_-<0<x_+$ and, moreover, \begin{equation*} x_\pm=\pm\Big(\frac c{d_n}\Big)^{1/n}-(1+o(1))\frac{d_{n-1}}{nd_n} \tag{1} \end{equation*} uniformly over all polynomials $p(x)=\sum_{j=0}^n d_j x^j$ in $\mathcal P_{n,d_*,D}$; here and in the sequel the asymptotic relations are for $$c\to\infty,$$ unless otherwise specified. This uniformity can be obtained by refining this reasoning.
Moreover, without loss of generality (wlog), \begin{equation*} \text{for all $p\in\mathcal P_{n,d_*,D}$ and all real $c\ge c_*(n,d_*,D)$ we have $p'(x_\pm)\ne0$.} \tag{1.5} \end{equation*} Indeed, because (1) holds uniformly over all $p\in\mathcal P_{n,d_*,D}$, wlog \begin{equation*} |x_\pm|\ge\Big(\frac cD\Big)^{1/n}-2\frac D{nd_*}\to\infty, \tag{1.6} \end{equation*} so that $|x_\pm|\to\infty$ uniformly over all $p\in\mathcal P_{n,d_*,D}$. Also, taking any polynomial $p(x)=\sum_{j=0}^n d_j x^j$ in $\mathcal P_{n,d_*,D}$ and writing $p'(x)=\sum_{j=1}^n d_j jx^{j-1}$, we see that for $|x|\ge1$ \begin{equation*} \frac{|p'(x)|}{|x|^{n-1}}\ge nd_n-\sum_{j=1}^{n-1} |d_j| j|x|^{j-n} \ge nd_*-n D |x|^{-1}\underset{x\to\infty}\longrightarrow nd_*>0. \end{equation*} So, by (1.6), wlog (1.5) holds indeed.