Relative extrema of multivariables: f(x,y)= (xy)/7 find critical points and relative extrema, given an open region.

We have, ${\displaystyle f(x,y)=\frac{xy}{7}}$
To find the critical point:
Find the partial derivative and equate to zero:
Now,
Differntiate f(x,y) with respect to x and equate to zero
${\displaystyle f_x(x,y)=\frac{y}{7}=0}$
${\displaystyle \Rightarrow y=0}$
Differntiate f(x,y) with respect to y and equate to zero
${\displaystyle f_y(x,y)=\frac{x}{7}=0}$
${\displaystyle \Rightarrow x=0}$
Hence, the critical point is (0,0)
Now we have to find the relative extrema.
Use:
${\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-{\left(f_{xy}(x,y)\right)}^2}$
Find D(0,0):
Now
${\displaystyle f_{xx}(0,0)=0}$
${\displaystyle f_{yy}(0,0)=0}$
${\displaystyle f_{xy}(0,0)=\frac{1}{7}}$
Therefore,
${\displaystyle D(0,0)=0\times 0-{\left(\frac{1}{7}\right)}^2=\frac{1}{49}<0}$
Hence, by the ${\displaystyle 2^{nd}}$ derivative test:
f(x,y) neither max. nor min at (0,0) which means (0,0) is saddle point.