I Finding Max and Min Extremes of a Function with Second Derivatives Equal to Zero

NODARman
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What should I do when the f(x, y) function's second derivatives or Δ=AC-B² is zero? When the function is f(x) then we can differentiate it until it won't be a zero, but if z = some x and y then can I just continue this process to find what max and min (extremes) it has?

What I've done is:
Differentiated z=f(x, y) by partial derivatives with respect to x and y;
Made them equal to zero in the system of numbers;
Where I got x=0, y=0;
Differentiated function by the second partial derivatives;
Used Δ = AC-B² where if Δ=0 then it needs more calculations.
I got Δ=0.

What should I do now? Differentiate again (third partial) as we do to function f(x) with only one variable x until it won't be equal to zero?
 
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What the determinant of the hessian matrix of f vanishing tells you is that is has a zero eigenvalue with a corresponding eigenvector \mathbf{v}. To determine the behaviour of f in this direction, you should look at <br /> \frac{d^3}{dt^3}f(\mathbf{x}_0 + t\mathbf{v}) and higher derivatives until you find one which does not vanish. If both eigenvalues are zero then the hessian is zero, and you need to look at the cubic terms to see what is going on.

Switching to a different coordinate system may give better insight into what is going on, for example https://www.wolframalpha.com/input?i=(x^2+y^2)*(1+-+(x^2+y^2)) where in polar coordinates it is easy to find that r^2(1-r^2) has a ring of global maxima at r = 1/\sqrt{2} and a local minimum at r = 0.
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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