Beneath i have the question, what i did and where iam stuck. (sorry for the formatting)
Given the following equation: $f(x, y) = x^4 − 6(x^2)(y^2)+ y^4$
With the question:
What is the nature of the critical point(s), max, min, saddle...? If the second order derivative doesn't give an answer use f[x, x] en f[x, 0].
I've taken the gradient of f(x,y);
$$df(x,y)/dx = 4x^3 - 12xy^2$$ $$df(x,y)/dy = 4y^3 - 12x^2y$$
I found the critical point (0,0). However if i now use the second derivative i get 0.
$$f_{xx}f_{yy} - f_{xy}^2 = 0$$
I now use f[x, x] en f[x, 0]?
$$f[x,x] = -4x^4$$ $$f[x,0] = x^4$$
But if i use the second derivative test here again i get 0 for both?
$$f''(0) = -48 * (0)^2 = 0$$ $$f''(0) = 12 * (0)^2 = 0$$
So now what? I read somewhere that if you have the fourth derivative you get an inflection point?
This is a case where, by design, the second derivative test fails.
As you have observed, the function has a minimum at the origin when restricted to the line $y=0$ and a maximum when restricted to a line $y=x$. Therefore it has neither a maximum, nor a minimum at the origin. It must have a saddle point of some sort, but not necessarily one of the usual shape.
It's convenient to make a polar substitution. Set $x=r\cos t$ and $y=r\sin t$. Then $$f(r\cos t,r\sin t)=r^4(\cos^4t-6\cos^2t\sin^2t+\sin^4t)=r^4\cos 4t.$$ If we take a circle centred at the origin and of radius $r$ then $f$ has four (equally spaced) maxima and four minima on the circle. Compare with a standard saddle point with two maxima and two minima on a small circle about that point. So it's a "saddle" but with two lines of "maximal ascent", namely $x=0$ and $y=0$ and two lines of "maximal descent": $y=x$ and $y=-x$, instead of one each.