How to determine if $z(x,y)=\ln(x^2 + y^2)$ is a harmonic function So I know that a function is harmonic if it satisfies Laplace's equation:

$\frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2} = 0$

But I'm just not sure how I should put it in to action for a practice question I have, which is:
Determine if $z(x,y) = \ln(x^2 + y^2)$ is a harmonic function.
By the Chain Rule, $\partial z/\partial x=2x/(x^2+y^2)$. Now by the quotient rule, $$\partial^2 z/\partial x^2=\frac{2(x^2+y^2)-2x(2x)}{(x^2+y^2)^2}=\frac{2(y^2-x^2)}{(x^2+y^2)^2}$$

Similarly $\partial^2 z/\partial y^2=\frac{2(x^2-y^2)}{(x^2+y^2)^2}$. But this is the negative of the previous answer, so they add to zero.
Any real function $u(x,y)$ with continuous second partial derivatives which satisfies Laplace's equation,

$$\nabla ^2u(x,y)=0$$is called a harmonic function.

$z(x,y)=\ln(x^2+y^2)$
$$\nabla ^2 z(x,y)=\frac{\partial^2 z}{\partial x^2}+\frac{\partial^2 z}{\partial y^2}$$
 you can solve the above expression and it must equal to $0$ to be satisfied for  the harmonic function