Q:

Show $f(x,y) = y^2 - x^2$ at $(0,0)$ has a critical point, but is not a max/min value

So as always... I found the partial derivative with respect to $x$ and $y$ of $f(x,y)$ which gave me:

$f_x=-2x$
$f_y=2y$

So I wasn't too sure what to do next, but I set $f_x = 0$:
$0 = -2x$
$x=0$
And I got stuck again. How do I continue AND prove that $f(x,y)$ at $(0,0)$ has a critical point but is not max/min value? Thanks!

A:

Intuitively: 
If $x=0$, then $f(y)=y^2$ has a minimum point at $y=0$; 
If $y=0$, then $f(x)=-x^2$ has a maximum point at $x=0$;
Thus, if $(x,y)=(0,0)$, then $f(x,y)=y^2-x^2$ has a saddle point. 

