I know if Hessian determinant is zero then second derivative test fails.. But I find that the following proof is a contradiction !
Moving along the unit vector $$u=u_1 i + u_2 j$$ and knowing that the crtical point is $$(x_0,y_0)$$
$$$$ then
$$F(h)=f(x_0+h u_1,y_o+h u_2)$$
Directional derivative in u direction$$F'(h)=D_u f=f_x u_1 + f_y u_2 $$
$$F''(h)=D_u^2f=f_{xx} u_1^2 +f_{yy} u_2^2 + 2 f_{xy} u_1 u_2$$
$$ F''(0)=f_{xx}(x_0,y_0)(u_1+\frac{f_{xy}(x_0,y_0)}{f_{xx}(x_0,y_0)} u_2)^2 +\frac{u_2^2}{ f_{xx}(x_0,y_0)}{(f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-f_{xy}^2(x_0,y_0)})$$
if $$ f_{xx}(x_0,y_0) >0 $$ and $$f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-f_{xy}^2(x_0,y_0) >0$$
then it is local minimum
$$ $$ if $$ f_{xx}(x_0,y_0) <0 $$ and $$f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-f_{xy}^2(x_0,y_0) >0$$
then it is local maximum
$$ $$if $$f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-f_{xy}^2(x_0,y_0) <0$$
then it is saddle point
$$ $$ if $$f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-f_{xy}^2(x_0,y_0) =0$$
then
$$ F''(0)=f_{xx}(x_0,y_0)(u_1+\frac{f_{xy}(x_0,y_0)}{f_{xx}(x_0,y_0)} u_2)^2 $$
I see that if the hessian determinant is zero , we did not fail to classify the point ! ..the point may be local max or min depending on the sign of $$f_{xx}$$
