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There are plenty of such polynomials and/or analytic functions, it suffices to perturb $x^2+y^2$ by adding a polynomial/function $g(x,y)$ whose Taylor expansion in zero has order at least 3. For instance, take $g(x,y)$ a polynomial whose monomials have degree at least 3. As an example, $g(x,y) = x^3$. You only need to care that $g$ is not divisible by $x^2+y^2$.

Concerning you remark, any continuous real function $f$ having an isolated zero on $(0,0)$ is nonnegative or nonpositive in a small neighbourhood of $(0,0)$. Since the zero is isolated, there is a neighborhood $U$ of $(0,0)$ such that $f(U\setminus{(0,0)})\subset \mathbb R$ does not contain 0. Since $U\setminus {(0,0)}$ is connected, its image $f(U\setminus {(0,0)})$ also is, and is therefore an interval in $\mathbb R$. An interval that does not contain zero consists of either all only positive or all only negative real numbers.
show/hide this revision's text 2 minor edit

There are plenty of such polynomials and/or analytic functions, it suffices to perturb $x^2+y^2$ by adding a polynomial/function $g(x,y)$ whose Taylor expansion in zero has order at least 3. For instance, take $g(x,y)$ a polynomial whose monomial monomials have degree at least 3. As an example, $g(x,y) = x^3$. You only need to care that $g$ is not divisible by $x^2+y^2$.

Concerning you remark, any continuous real function $f$ having an isolated zero on $(0,0)$ is nonnegative or nonpositive in a small neighbourhood of $(0,0)$. Since the zero is isolated, there is a neighborhood $U$ of $(0,0)$ such that $f(U\setminus{(0,0)})\subset \mathbb R$ does not contain 0. Since $U\setminus {(0,0)}$ is connected, its image $f(U\setminus {(0,0)})$ also is, and is therefore an interval in $\mathbb R$. An interval that does not contain zero consists of either all positive or all negative real numbers.
show/hide this revision's text 1

There are plenty of such polynomials and/or analytic functions, it suffices to perturb $x^2+y^2$ by adding a polynomial/function $g(x,y)$ whose Taylor expansion in zero has order at least 3. For instance, take $g(x,y)$ a polynomial whose monomial have degree at least 3. You only need to care that $g$ is not divisible by $x^2+y^2$.

Concerning you remark, any continuous function $f$ having an isolated zero on $(0,0)$ is nonnegative or nonpositive in a small neighbourhood of $(0,0)$. Since the zero is isolated, there is a neighborhood $U$ of $(0,0)$ such that $f(U\setminus{(0,0)})\subset \mathbb R$ does not contain 0. Since $U\setminus {(0,0)}$ is connected, its image $f(U\setminus {(0,0)})$ also is, and is therefore an interval in $\mathbb R$. An interval that does not contain zero consists of either all positive or all negative real numbers.