Discuss maxima/minima of the following function: $$f(x,y)=2(x^4+y^4+1)-(x+y)^2$$
It's easy to compute the gradient and the hessian of the function: $$\nabla(f)=(8x^3-2(x+y),8y^3-2(x+y)) \qquad H(f)=\begin{bmatrix}24x^2-2 & -2 \\ -2 & 24y^2-2\end{bmatrix}$$ from which we can easily see that $\nabla(f)=0 \iff (x,y)\in\{(0,0),(\pm \frac{1}{\sqrt2},\pm\frac{1}{\sqrt2})\}$ and by studing the hessian, we can understand that $f$ has a minima in $P_{1/2}(\pm \frac{1}{\sqrt2},\pm\frac{1}{\sqrt2})$.
The real problem is to discuss what happens in $(0,0)$. Clearly we have that $f(0,0)=2$ but the second derivative test fails as the determinant of $H(f)(0,0)$ is $0$. How can I proceed?
You can try to use Taylor's formula and study the sign of the higher order differentials. However, it is easier to study directly what happens around $(0,0)$.
If you restrict yourself to the sets $B_m=\{(x,y): y=mx\}$ and study the function $\phi_m(x) = f(x,mx)=2(1+m^4)x^4+2-(1+m)^2x^2$, you can observe that,
$$ \phi'(0) = 0, \quad \phi''(0) = -2(1+m)^2\leq 0. $$
So, if $m \ne 1$, $\phi_m$ attains a maximum at $x=0$.
If $m=-1$, $$\phi_{-1}''(0)=0, \phi_{-1}'''(0) = 0, \phi_{-1}^{(4)}(0) = 96 > 0,$$ which means that $\phi_{-1}$ attains a minimum at $x_0$.
Considering these two facts, in every neighbouhood of $(0,0)$ there are points $(x,y)$ for which $f(x,y)>f(0,0)$ (the ones in $B_{-1}$) and points $(x,y)$ for which $f(x,y)<f(0,0)$ (for instance the ones in $B_1$). The conclusion is that $(0,0)$ is a saddle point.