Help please, find the domain of f(x, y)= sqrt( |x|+|y|-3 ). Thanks?

The domain is all points that satisfy the equation

.. |x| + |y| ≥ 3

This defines the area on and outside the rhombus whose vertices are ±3 on the x- and y-axes. See the source link for a plot.

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It will be the union of the areas ...

.. (x+y ≥ 3) U (x+y ≤ -3) U (x-y ≥ 3) U (x-y ≤ -3)

or

.. (y ≥ x+3) U (y ≥ -x+3) U (y ≤ x-3) U (y ≤ -x-3)

I'm delighted to see that the answer given above by "Anonymous" is essentially the same answer I gave you several hours ago. If you are dealing with a machine, you're going to have a hard time figuring out what format IT wants for answer.

The existence of the domain requires |x|+|y|-3>=0

=>[y>=-x+3 or y>=x+3]or[y<=x-3 or y<=-x-3]

=>domain={(x,y)| y>=-x+3}U{(x,y)| y>=x+3}U{(x,y)| y<=x-3}U{(x,y)| y<=-x-3}

=>domain=the whole x-y plane except the closed region formed by

(3,0), (0,3), (-3,0) & (0,-3) 4 points; the square.

The domain is an open region.

All point (x,y) of the plane that are on or outside of the square whose vertices are +3, -3 on x-axis and +3, -3 on y-axis.