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A level curve is the planar level set of a scalar function. Where the gradient does not vanish, the implicit function theorem makes it locally a smooth curve. At a critical point, branches may cross, narrow into a cusp or shrink to a point, so a plot should be interpreted with the local critical-point analysis.

Expand the function into separated terms: $f=g(x)+h(y)$, where $g=x^2-x^4$ and $h=y^8-y^4$. ThenPut $a=2^{-1/2}$ and $b=2^{-1/4}$. The nine critical points are all pairs from $x\in \{0,\pm a\}$ and $y\in \{0,\pm b\}$. The diagonal Hessian matrix has entries $2-12x^2$ and $56y^6-12y^2$.

At $(0,\pm b)$ both entries are positive, giving strict local minima of value $-1/4$. At $(\pm a,\pm b)$ they have opposite signs, giving four saddle points of value zero. At the three points on $y=0$ the Hessian is degenerate, so its sign alone does not classify them. Use the higher-order test for separated extrema: $h(y)=-y^4+O(y^8)$ is negative for small nonzero $y$. At $(0,0)$ the positive $x^2$ term and negative $y^4$ term give a saddle. At $(\pm a,0)$, $g-1/4=-(x^2-1/2{)}^2$, so every sufficiently small nonzero displacement decreases $f$ and these are strict local maxima.

For the level curves, the exact zero set consists of the two parabolas $x=\pm y^2$ and the oval $x^2+y^4=1$. They cross at the four nondegenerate saddle points. Small positive levels enclose each maximum; small negative levels enclose each minimum. The degenerate central saddle has quartic narrowing, locally $x^2-y^4=c$. Far along the $x$ axis the function is negative, while far along the $y$ axis it is positive. These signs and zero curves determine the connectivity shown in the sketch.