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.

The surface you are describing can be described as $xyz = 1$.  Since this is symmetric in $x$ versus $y$ versus $z$, any point where $|x| = |y| = |z|$ would be an extremum in distance from the origin.  Visually, we can see that that such points are minima.
Therefore the solutions must be $x,y,z = (1,1,1), (1,-1,-1), (-1,1,-1),$ or $(-1,-1,1)$.  Only one of these has $x,y > 0$.
