Originally Posted by
HallsofIvy Okay, what have YOU done? Since this function is continuous for all x and y, a "critical point" is any point where the partial derivatives are 0. What are the partial derivatives of f?
"On any line through the origin", y= mx for some number m. Replace y with mx in f(x,y) so you have f(x,mx), a function of the single variable x. Show that that function has a local minumum at (0,0).
(Well, almost "any line through the origin". The vertical line through the origin is x=0 and cannot be written that way. Look at f(0,y)= y^2.)