Is this function harmonic?

Question

$$u(x,y)=\frac{x}{x^2+y^2}$$

I'm trying to figure out if this function is harmonic. I worked out all the algebra and my conclusion is no because the numerator of the final fraction is $2x^3+2xy^2$ which doesn't cancel to zero. I did this by finding the second derivatives of both $x$ and $y$ and adding them together. Does anyone know if this function is supposed to be harmonic? I don't know the syntax for Wolfram for this. Anyway, below is my final answer:

$$=\frac{2x^3+2xy^2}{(x^2+y^2)^3}$$

Answer 1

Pedantic answer: no, it's not harmonic at $(0,0)$, since it's not defined there.

Useful answer: what about away from $(0,0)$? $$ \partial_y^2 \frac{x}{x^2+y^2} = -2x\partial_y \frac{y}{(x^2+y^2)^2} = -2x\frac{x^2+y^2-4y(x^2+y^2)^2}{(x^2+y^2)^3} = \frac{8 x y^2}{\left(x^2+y^2\right)^3}-\frac{2 x}{\left(x^2+y^2\right)^2}, $$ and similarly, $$ \partial_x^2 \frac{x}{x^2+y^2} = \dotsb = \frac{8 x^3}{\left(x^2+y^2\right)^3}-\frac{6 x}{\left(x^2+y^2\right)^2}, $$ and adding gives zero.

Clever answer: with $z=x+iy$ we have $$ \frac{1}{z} = \frac{1}{x+iy} = \frac{x-iy}{(x+iy)(x-iy)} = \frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}, $$ so away from $(0,0)$, your function is the real part of the analytic function $1/z$, and hence is harmonic.

Answer 2

Wolfram can do this.

Without showing work it's hard to say where your calculation was incorrect. A quick way to see that your function is harmonic is that it is the real part of the holomorphic function (away from $0$) $\frac{1}{z}$.