Point on $z = \frac{1}{xy}$ closest to origin Where $x>0$ and $y>0$.
I want to work with the square of the distance formula from the origin, so I went with $f(x,y) = x^2 + y^2 + \frac{1}{(xy)^2}$.
Then I found the first partial derivatives:
$f_x = 2x - \frac{2}{x^3y^2}$ and $f_y = 2y - \frac{2}{y^3x^2}$.
Set them equal to $0$ to find the critical points, and multiply both sides by respective $x$ and $y$ to get a common term to set equal to each other:
$2xy = \frac{2}{x^3y}$ and $2xy = \frac{2}{y^3x}$ so $\frac{2}{x^3y} = \frac{2}{y^3x}$. Simplfying, we get $y^2 = x^2$, or $y = x$ since we are given that $x>0$ and $y>0$.
I'm not sure what to do from here.

Alternative solution: The Inequality of arithmetic and geometric means 
(with $n=3$) states that for non-negative real numbers $a, b, c$
$$
      \frac{a + b + c}3 \ge (abc)^{1/3}
$$
with equality if and only if $a=b=c$. It follows that if $(x, y, z)$ is a point
on the surface $z = 1/(xy)$ then
$$
     \frac{x^2 + y^2 + z^2}3 \ge (x^2y^2z^2)^{1/3} = 1
$$
and therefore $x^2 + y^2 + z^2 >= 3$ with equality if and only if $x=y=z$ ($= 1$).
