This is a proof which only uses directional derivatives.

We want to prove the Second Derivative Test.

Suppose the second partial derivatives of \(f\) are continuous on a disk with center (a, b), and suppose that \(f_x(a,b)=f_y(a,b)=0\). Let

\[H=\begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix}=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2\]

(a) If \(H>0, f_{xx}(a,b)>0\), then \(f(a,b)\) is a local minimum.

(b) If \(H>0, f_{xx}(a,b)<0\), then \(f(a,b)\) is a local maximum.

(c) If \(H<0\), then \(f(a,b)\) is a saddle point.

(d) If \(H=0\), inconclusive.

Let \(u=(h, k)\), which is a unit vector.

\[\begin{align} D^2_uf(x,y)&=D_u[D_uf(x,y)]\\ &=D_u[\nabla f\cdot u]\\ &=(f_{xx}h+f_{yx}k)h+(f_{xy}h+f_{yy}k)k \end{align}\]

Since the second partial derivatives are continuous, \(f_{xy}=f_{yx}\)

\[\begin{align} D^2_uf(x,y)&=f_{xx}h^2+2f_{xy}hk+f_{yy}k^2\\ &=k^2(f_{xx}(\frac{h}{k})^2+2f_{xy}\frac{h}{k}+f_{yy}) \end{align}\]

Let \(z=\frac{h}{k}\), \(g(z)=f_{xx}z^2+2f_{xy}z+f_{yy}\)

Use the quadratic function above to find the range of solutions of \(D^2_uf(x,y)\).

Let \(d=(2f_{xy})^2-4f_{xx}f_{yy}=4(f_{xy}^2-f_{xx}f_{yy})=-4H\)

(1) If \(d<0\), means that \(D^2_uf(x,y)\) can only be either positive or negative.

We can know that the point \((a,b)\) must be a local minimum or a local maximum.

\[\begin{align} D^2_uf(x,y)&=f_{xx}h^2+2f_{xy}hk+f_{yy}k^2\\ &=f_{xx}(h^2+2\frac{f_{xy}}{f_{xx}}hk+\frac{f_{yy}}{f_{xx}}k^2)\\ &=f_{xx}[(h+\frac{f_{xy}}{f_{xx}}k)^2+(f_{xx}f_{yy}-f_{xy}^2)k^2]\\ &=f_{xx}[(h+\frac{f_{xy}}{f_{xx}}k)^2+Hk^2] \end{align}\]

We know that \(H>0\), so,

if \(f_{xx}>0\), the equation above will always be positive, which means a local minimum at \((a,b)\).

Otherwise if \(f_{xx}<0\), it will always be negative, which means a local maximum at \((a,b)\).

(2) If \(d>0\), means that \(D^2_uf(x,y)\) can be both positive, zero or negative, so

\(H<0\) tells us that the function has a saddle point at \((a,b)\).

(3) If \(d>0\), means that \(D^2_uf(x,y)\) can only be either positive and zero or negative and zero, so

\(H=0\) cannot tell that whether \((a,b)\) is a local minimum, local maximum or a saddle point.

Note: Posted at The “second derivative test” for $f(x,y)$.