Solutions of the Laplace equation How do I find solutions $u=f(r)$ of the two-dimensional Laplace equation $u_{xx}+u_{yy}=0$ that depend only on the radial coordinate $r= \sqrt{x^2+y^2}$

As @CameronWilliams pointed out, the easiest way to do this is to switch to polar coordinates.    But, here is a brute force way to proceed.  
Let $f(x,y)=g(x^2+y^2)$.  Then, $\frac{\partial f}{\partial x}=2xg'(x^2+y^2)$.
Taking a second partial yields $\frac{\partial^2 f}{\partial x^2}=(2x)^2g''(x^2+y^2)+2g'(x^2+y^2)$.
One can easily see then (replace $x$ with $y$) that $\frac{\partial^2 f}{\partial y^2}=(2y)^2g''(x^2+y^2)+2g'(x^2+y^2)$.
Thus, 
$$\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=4(x^2+y^2)g''(x^2+y^2) + 4g'(x^2+y^2)=0$$
Then, we may write 
$$\frac{g''(x^2+y^2)}{g'(x^2+y^2)}=\left(\log g'(x^2+y^2)\right)'=-\frac{1}{x^2+y^2}$$
where again the "prime" is on $x^2+y^2$.  Integrating once reveals that $\log g'(x^2+y^2) = -\log(x^2+y^2) +C \Rightarrow g(x^2+y^2) = A\log (x^2+y^2) +B$.
