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?
 A: As the OP has answered the question himself, I just want to add the general solution of this PDE can be obtained via the method of characteristics. For that one converts the PDE in an ODE along the characteristic line. The system of ODE which needs to be solved reads
$$\frac{dx}{ds} = x, \frac{dy}{ds} = y, \frac{df}{ds}= \frac{1}{r}; $$
where $s$ parameterizes the characteristic line.
The general solution for $x$ and $y$ is given by
$$\tag{1}x= x_0 e^{s},y=y_0 e^{s};$$
thus we have $$\frac{df}{ds} = \frac{e^{-s}}{\sqrt{x_0^2+y_0^2}} $$
which is solved by
$$f= f_0(y_0/x_0) - \frac{e^{-s}}{\sqrt{x_0^2+y_0^2}}$$
where $f_0$ depends on the characteristic  line via $y_0/x_0$.
Using (1) to eliminate $x_0$ and $y_0$, we obtain
$$f= f_0(y/x) - \frac{1}{\sqrt{x^2+y^2}}$$
with an arbitrary function $f_0$.
A: 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: $g(r)=C-\frac{1}{\sqrt{x^2+y^2}}$
