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As in the title I found critical points to be $(1,0)$-saddle point, $(-1,0)$- local minimum and $(0,0)$ is the one I have problem with. Second derivative test is inconclusive in this case. If it was a saddle point I should find some curve in the graph that has inflection point at $(0,0)$ to prove it. Unfortunately, I suspect it to be local maximum and in this case I have no clue how can I prove that it grows in neighborhood of $(0,0)$ in every direction.

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    $\begingroup$ The function can be rewritten in the form $$f(x,y)=x^4+2x^2-(y^2+2x)^2.$$ Can you use that?! $\endgroup$ – Jyrki Lahtonen Jun 30 at 20:59
  • $\begingroup$ Aren't the (real) critical points $(\pm 1,0)$ and $(0,0)$? $\endgroup$ – uniquesolution Jun 30 at 21:01
  • $\begingroup$ @AjayMishra I disagree. We see that along the curve $y^2=-2x$ through the origin $f$ takes positive values. OTOH along the the curve $x=0$ through the origin $f$ takes negative values. Therefore... $\endgroup$ – Jyrki Lahtonen Jul 1 at 4:36
  • $\begingroup$ @JyrkiLahtonen yeah thanks. That’s neat way to show it is a saddle $\endgroup$ – Kran Jul 1 at 9:29
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Consider the following graphs: \begin{align} y = x & \implies x^4-y^4-4xy^2-2x^2= -4x^3-2x^2\\ &\implies x=0\text{ is a local maximum along the curve}\\ y^2 = (-2\pm \sqrt{2})x, x<0 & \implies x^4-y^4-4xy^2-2x^2= x^4\\ &\implies x=0\text{ is a local minimum along the curve} \end{align} Hence, $(0,0)$ is a saddle point.

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  • $\begingroup$ A point at which derivative vanishes, but is not a local extremum. Here, the function value increases along one direction, and decreases along the other. Hence (0,0) is a saddle point. Wikipedia:en.m.wikipedia.org/wiki/Saddle_point $\endgroup$ – Geethu Joseph Jul 1 at 4:15
  • $\begingroup$ Yes. $x^3$ increases as we move along the positive direction and decreases as we move along the negative direction. To prove that a point is saddle it is sufficient to find two directions such that function increases along one and decreases along the other. $\endgroup$ – Geethu Joseph Jul 1 at 4:23
  • $\begingroup$ Definition 2.2 says that the two curves should transversly intersect, which doesn't mean that they are orthogonal. Clearly, the tangent space of the graph $y=x$ at (0,0) is the same curve. The tangent space of the other graph at (0,0) is not along the curve $y=x$. So you have a tangent vector on the curve $y=x$ (which is along the curve $y=x$) and a tangent vector on the other curve (which is not along the curve $y=x$). Hence these vectors are linearly independent, and span the whole plane. Thus, they intersect transversly. en.m.wikipedia.org/wiki/Transversality_(mathematics) $\endgroup$ – Geethu Joseph Jul 1 at 4:57
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Here, I am following this definition of saddle points.

For the given function, $$f(x,y) = x^4 - y^4 - 4xy^2 -2x^2$$ $\nabla f = \vec 0$ , so it is a stationary point. Now looking at the level surfaces when $f(x,y) = 0$ $$\implies x^4 - y^4 - 4xy^2 - 2xy^2 = 0$$ $$\implies y = \pm \sqrt{-2x \pm x \sqrt{x^2 + 2}}$$ enter image description here Whose implication is that that along these curve, value of $f(x,y) = 0$ remains the same, so following the definition of extremes, it should be obvious to conclude they are not one of them, hence they must be saddle point. Here is the graph of the function in the vicinity of $(0,0,0)$ enter image description here

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