Schwarz–Clairaut symmetry of mixed partial derivatives and Hessian test

For small nonzero $h, k$, consider the second difference $\Delta(h,k) = f(x_0+h,y_0+k) - f(x_0+h,y_0) - f(x_0,y_0+k) + f(x_0,y_0)$. Applying the one-variable Mean Value Theorem first to $g(x) = f(x,y_0+k) - f(x,y_0)$ on $[x_0, x_0+h]$ and then to $f_x$ along $y$ yields $\Delta(h,k) = hk\,f_{yx}(c_1, d_1)$, while applying it in the reverse order yields $\Delta(h,k) = hk\,f_{xy}(c_2, d_2)$ for intermediate points converging to $(x_0,y_0)$. Equating the two and letting $(h,k) \to (0,0)$ proves $\frac{\partial^2 f}{\partial x\partial y}(x_0,y_0) = \frac{\partial^2 f}{\partial y\partial x}(x_0,y_0)$ by continuity.

At a critical point where $\nabla f(x_0,y_0) = \mathbf{0}$, Taylor's formula gives $f(x_0+h,y_0+k) - f(x_0,y_0) = \frac{1}{2}Q(h,k) + o(h^2+k^2)$, where $Q(h,k) = f_{xx}h^2 + 2f_{xy}hk + f_{yy}k^2$. When $f_{xx} \ne 0$, completing the square writes $Q(h,k) = \frac{1}{f_{xx}}\left[(f_{xx}h + f_{xy}k)^2 + Dk^2\right]$ with $D = f_{xx}f_{yy} - f_{xy}^2$. If $D > 0$, the bracketed sum is strictly positive for all $(h,k) \ne (0,0)$, so the sign of $Q(h,k)$ matches the sign of $f_{xx}$ (giving a strict minimum for $f_{xx} > 0$ and maximum for $f_{xx} < 0$). If $D < 0$, $Q(h,k)$ takes both positive and negative values along different lines through the origin, producing a saddle point.