I'll take a guess about what you're trying to do.
As you write, you've got a scalar function of two variables. Since you say you're trying to calculate a surface area, we have to somehow interpret this as defining a surface. A straightforward way to do that would be to take this function as the $z$ component of a surface over the $x$-$y$ plane.
We can treat this surface just like any parameterized surface by regarding $x$ and $y$ as parameters which happen to coincide with two of the coordinates. Then the parametrization of the surface is
$\vec r (x,y) = (x,y,f(x,y))=\left(x,y,\sqrt{x^2+y^2}\right)\;.$
Then the tangent vectors along the coordinate lines are
$\frac{\partial\vec r(x,y)}{\partial x}=\left(1,0,\frac x{\sqrt{x^2+y^2}}\right)$
and
$\frac{\partial\vec r(x,y)}{\partial y}=\left(0,1,\frac y{\sqrt{x^2+y^2}}\right)\;,$
and the surface element is
$ \begin{eqnarray} \mathrm dS &=& \left|\frac{\partial\vec r(x,y)}{\partial x}\times\frac{\partial\vec r(x,y)}{\partial y}\right|\,\mathrm dx \mathrm dy \\ &=& \left|\left(-\frac x{\sqrt{x^2+y^2}},-\frac y{\sqrt{x^2+y^2}},1\right)\right|\,\mathrm dx \mathrm dy \\ &=& \sqrt2\mathrm dx \mathrm dy\;. \end{eqnarray} $
What this tells you is that the surface element of a cone at an angle $45^\circ$ (which is the surface described by that function) is $\sqrt2$ (the reciprocal of the sine of the cone's angle) times the projected surface element in the $x$-$y$ plane.
I hope that's relevant to what you had in mind.