We have if $(a,b)$ is a critical Point, $f_{xx}\ne 0$ and:
$$f_{xx}f_{yy}-f^2_{xy}=0$$
By Taylor's Formula and using Operator notation we get:
$$f(a+h,b+k)=f(a,b)+\left(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y}\right)f(a,b)+\frac{1}{2!}\left(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y}\right)^2f(a,b)+....$$ So at the Critical point we have $f_x=f_y=0$
$$f(a+h,b+k)-f(a,b)=\frac{1}{2}\left(h^2f_{xx}+2hkf_{xy}+k^2f_{yy}\right)+R(h,k)$$
$$f(a+h,b+k)-f(a,b)=\frac{1}{2 f_{xx}}\left(\left(hf_{xx}+kf_{xy}\right)^2+k^2(f_{xx}f_{yy}-f^2_{xy})\right)+R(h,k)$$
That gives finally $$f(a+h,b+k)-f(a,b)=\frac{1}{2f_{xx}}\left(hf_{xx}+kf_{xy}\right)^2+R(h,k)$$
Now when $$hf_{xx}+kf_{xy}\ne 0$$ Then why cant we conclude as follows
Case $1.$ Local Maximum if $f_{xx}\lt 0$
Case $2.$ Local Min if $f_{xx} \gt 0$
Only if $$hf_{xx}+kf_{xy}=0$$ We should go for Higher derivatives
Am i missing something here?
The point is that to say that $(a,b)$ is a local minimum point, say, you need to know that $f(a+h,b+k)-f(a,b) \ge 0$ for all (small) values of $h$ and $k$. This includes those values for which $hf_{xx}+kf_{xy}=0$. So the test remains inconclusive.
Consider the examples $f(x,y) = x^2+y^3$, $f(x,y)=x^2+y^4$, and $f(x,y)=x^2-y^4$ at the origin. (Indeed, with $h=0$ and $k$ arbitrary, we have $hf_{xx}+kf_{xy}=0$, and it's precisely along the $y$-axis that we need to study the function further.)