I am trying to prove a point is a saddle point, and the only information I have about it is that the value of the function is $0$ at this point and the determinant of the hessian is negative at the point. All the proofs I have found online and on this site use the eigenvalues of the hessian, however I have never learnt this. So far, I have wrote the second degree taylor expansion but am stuck from here. Any help is expected.
The question: “If $f: \mathbb{R^2}\to\mathbb{R}$ is a $C^3$ function such that $f(0,0)=0$ and $det(H(0,0))<0$. Prove that $(0,0)$ is a saddle point.
What you're trying to prove is false:
$$ f(x, y)= x^2 - y^2 + 52x $$ is a counterexample to the problem as stated; you really DO need for $(0,0)$ to be a critical point as well, as Robert Lewis stated.
Once you know that, it might help you to remember that the determinant of a $2 \times 2$ matrix is the product of its (two) eigenvalues.