Failure of the Second Partials Test (Multivariable Calculus): $f(x,y) = \sqrt{x^2+y^2}$ I am asked to find the relative extrema of $$f(x,y) = \sqrt{x^2+y^2}$$ 
The partials that I get are $$f_x=\frac{x}{\sqrt{x^2+y^2}}\\f_y=\frac{y}{\sqrt{x^2+y^2}}$$ 
and to find the extrema I must do
$$f_x=\frac{x}{\sqrt{x^2+y^2}}=0\\f_y=\frac{y}{\sqrt{x^2+y^2}}=0$$ 
on which I cannot say $x=y=0$ (or can I?)
How can I interpret the minimum on $f(0,0)$? As some point to which the partials do not exist?
 A: A change into polar coordinates would give:
$$f(r, \theta) = r $$
Then you can show $f$ has a relative minimum at $0$ for the interval $[0, \infty]$.
(I could not comment so wrote this as an answer).
A: 
How can I interpret the minimum on $f(0,0)$? As some point to which the partials do not exist?

The fact that the partial derivatives don't exist, doesn't mean there is no extremum. Compare with the single-variable real-valued function $f : \mathbb{R}^+ \to \mathbb{R} : x \mapsto \sqrt{x}$ which doesn't have a (finite) derivative at $x=0$, but clearly has a minimum there.
That being said, you could also see $\sqrt{x^2+y^2}$ as the composition $\sqrt{u}$ with $u=x^2+y^2$, minimal at $u=0$ so for $x^2+y^2=0$, which only happens at $(x,y)=(0,0)$.
A: x^2+y^2=r^2  → y= ± (r^2-x^2)^0.5 is a circle; the derivative of y on x
 is
 -df(x, y)/dx)/df(x,y)/dy= - f'(x)/f'(y)=-x/y .
For finding extremes -x/y = 0 therefore x=0 , hence extremes(maximum and minimum) are y= ±r. if the derivative on y is considered then f'(y)/f'(x)=y/x=0 →  y=0
→  x=±r are extremes. 
Notice that f'(x) =2x is derivative of f(x, y) on x and f'(y)= 2y is the derivative of f(x, y) on y and y=(r^2 -x^2)^0.5 therefore derivative of y=(r^2-x^2)^0.5 is y'= -f'(x)/f'(y)=-2x/2y=-x/y  . So ther is no point like(0, 0) but there are points like(0,± r) and ( ±r, y) as extremes of the function.       
