Partial differential equation with chain rule Problem statement: 
Consider the PDE: $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=\frac{1}{\sqrt{x^2+y^2}}, (x,y)\neq (0,0) $
Determine all solutions to the equation of the form
$f(x,y)=g(r)$ where $r= \sqrt{x^2+y^2}$ 
My progress:
Now $f(x,y)=g(r)=g(\sqrt{x^2+y^2})$ and so the partial derivatives are:
$\frac{\partial f}{\partial x}=g'(\sqrt{x^2+y^2})\frac{x}{\sqrt{x^2+y^2}}$
$\frac{\partial f}{\partial y}=g'(\sqrt{x^2+y^2})\frac{y}{\sqrt{x^2+y^2}}$
Now our PDE will become:
$\frac{x^2}{\sqrt{x^2+y^2}}+\frac{y^2}{\sqrt{x^2+y^2}}=\frac{1}{\sqrt{x^2+y^2}} $
And furthermore: $x^2+y^2=1$ which is an equation for a circle at origo. So how did I end up wrong?

I actually solved it myself:
$\frac{\partial f}{\partial x}=g'(r)\frac{x}{r}$ and
$\frac{\partial f}{\partial y}=g'(r)\frac{y}{r}$
This gives: $g'(r)\frac{x^2+y^2}{r}=\frac{1}{r} \Rightarrow rg'(r)=\frac{1}{r}$
With solution: $g(r)=C-\frac{1}{\sqrt{x^2+y^2}}$
