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Recent addition: Inspired by the comments by DimaPasechnik and Matt F. about sum of squares decompositions, I tried the following very natural idea: Try to find $f$ of the form $f(x,y)=A(x,y)^2+B(x,y)^2$, where $A$ and $B$ are quadratic polynomials whose zero sets intersect in $4$ points, which then of course are local minima, and attempt to get a fifth local minimum. By a shift we may look for it in $(0,0)$. And indeed, it is not hard to arrange things which yields an example which can be checked almost effortlessly. One possibility is \begin{align} A &= 4y^2-1\\ B &= 9y^2 + 4y - x^2. \end{align} The curves $A=0$ and $B=0$ intersect in the four points \begin{equation} (\pm\frac{1}{2}, -\frac{1}{2}), (\pm\frac{\sqrt{17}}{2}, \frac{1}{2}). \end{equation} The partial derivatives of $A^2$ and $B^2$ clearly vanish in $(0,0)$, so $(0,0)$ is a critical point of $f=A^2+B^2$. The Hesse matrix however is only positive semidefinite. Nevertheless, $(0,0)$ is a local isolated minimum, because $f(x,y)=1+8y^2+$ terms of degree $\ge3$ and $f(x,0)=x^4+1$.

A similar example, where the Hessian in $(0,0)$ actually is positive definite, is $f=A^2+B^2$ for $A=x^2 + y - 2$ and $B = 3x^2 - 2y^2 + 2y + 1$. Note that $f=5+2x^2 + y^2+$ terms of degree $\ge3$.

Recent addition: Inspired by DimaPasechnik's and Matt F.'s comments about sum of squares decompositions, I tried the following very natural idea: Try to find $f$ of the form $f(x,y)=A(x,y)^2+B(x,y)^2$, where $A$ and $B$ are relatively prime quadratic polynomials whose zero sets intersect in $4$ points, which then of course are local minima, and attempt to get a fifth local minimum. By a shift we may look for it in $(0,0)$. And indeed, there are examples which can be checked instantly by hand: One possibility is \begin{align} A &= x^2 - y - 6\\ B &= 8x^2 - y^2 - 3y + 2. \end{align} The curves $A=0$ and $B=0$ intersect in the four points \begin{equation} (\pm1, -5)\text{ and } (\pm4, 10). \end{equation} As \begin{equation} f(x, y) = 40 + 20x^2 + 6y^2 + \text{terms of degree }\ge3, \end{equation} we see that $(0, 0)$ is a fifth local minimum.

Original answer: The quartic \begin{equation} f = 3x^4 - 5x^2y^2 + 5y^4 + 20x^2y - 32y^3 + x^2 + 2y^2, \end{equation} which is even in $x$, has $5$ isolated local minima in \begin{equation} (0, 0), (\pm2, -1), (\pm2, 5). \end{equation} In order to prove the claim, one checks that the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ vanish and that the Hessian is positive definite in these points, where the latter is equivalent to the traces and the determinants being positive.

A similar example, where the Hessian in $(0,0)$ actually is positive definite, is $f=A^2+B^2$ for $A=x^2 + y - 2$ and $B = 3x^2 - 2y^2 + 2y + 1$. Note that $f=5+2x^2 + y^2+$ terms of degree $\ge3$.

Original answer: The quartic \begin{equation} f = 3x^4 - 5x^2y^2 + 5y^4 + 20x^2y - 32y^3 + x^2 + 2y^2, \end{equation} which is even in $x$, has $5$ isolated local minima in \begin{equation} (0, 0), (\pm2, -1), (\pm2, 5). \end{equation} In order to prove the claim, one checks that the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ vanish and that the Hessian is positive definite in these points, where the latter is equivalent to the traces and the determinants being positive.

Recent addition: Inspired by the comments by DimaPasechnik and Matt F. about sum of squares decompositions, I tried the following very natural idea: Try to find $f$ of the form $f(x,y)=A(x,y)^2+B(x,y)^2$, where $A$ and $B$ are quadratic polynomials whose zero sets intersect in $4$ points, which then of course are local minima, and attempt to get a fifth local minimum. By a shift we may look for it in $(0,0)$. And indeed, it is not hard to arrange things which yields an example which can be checked almost effortlessly. One possibility is \begin{align} A &= x^2-1\\ B &= 2x^2 + xy - y^2 + 2x. \end{align} The curves $A=0$ and $B=0$ intersect in the four points \begin{equation} (-1, 0), (-1, -1), (1, \frac{1\pm\sqrt{17}}{2}). \end{equation} The partial derivatives of $A^2$ and $B^2$ clearly vanish in $(0,0)$, so $(0,0)$ is a critical point of $f=A^2+B^2$. The Hesse matrix however is only positive semidefinite. Nevertheless, $(0,0)$ is a local isolated minimum, because $f(x,y)=1+2x^2+$ terms of degree $\ge3$ and $f(0,y)=y^4+1$.

Recent addition: Inspired by the comments by DimaPasechnik and Matt F. about sum of squares decompositions, I tried the following very natural idea: Try to find $f$ of the form $f(x,y)=A(x,y)^2+B(x,y)^2$, where $A$ and $B$ are quadratic polynomials whose zero sets intersect in $4$ points, which then of course are local minima, and attempt to get a fifth local minimum. By a shift we may look for it in $(0,0)$. And indeed, it is not hard to arrange things which yields an example which can be checked almost effortlessly. One possibility is \begin{align} A &= 4y^2-1\\ B &= 9y^2 + 4y - x^2. \end{align} The curves $A=0$ and $B=0$ intersect in the four points \begin{equation} (\pm\frac{1}{2}, -\frac{1}{2}), (\pm\frac{\sqrt{17}}{2}, \frac{1}{2}). \end{equation} The partial derivatives of $A^2$ and $B^2$ clearly vanish in $(0,0)$, so $(0,0)$ is a critical point of $f=A^2+B^2$. The Hesse matrix however is only positive semidefinite. Nevertheless, $(0,0)$ is a local isolated minimum, because $f(x,y)=1+8y^2+$ terms of degree $\ge3$ and $f(x,0)=x^4+1$.