Why doesn't $x^2+y^2-4xy$ have a local minimum at $(0,0)$?

Question

Consider $f(x,y)= x^2+y^2-4xy. \nabla f = (2x-4y, 2y-4x),$ which equals $0$ at $(0,0$). $f_{xx}=2,f_{yy} =2,f_{xy}=f_{yx} = -4.$ Since $f_{xx}f_{yy}<(f_{xy})^2 $, $(0,0)$ is only a saddle point. But since $f_x=f_y=0$ and $f_{xx}, f_{yy}>0$, from the point of view of single variable calculus, shouldn't $(0,0)$ be a local minimum?
I somehow find this aspect of multivariable calculus much less intuitive. I'd be grateful for any help on this.