show $\frac{y}{x^2+y^2} $ is harmonic except at $y=0,x=0$ Let $f(z)=u(x,y)+iv(x,y) $
where $$ f(z)=u(x,y)=\frac{y}{x^2+y^2}$$
show $u(x,y)$ is harmonic except at $z=0$

Attempt
$$ u=\frac{y}{x^2+y^2}=y(x^2+y^2)^{-1} $$
Partial derivatives with x
$$\begin{aligned} 
      u_x&= y *(x^2+y^2)^{-2}*-1*2x
   \\    &= -y*2x(x^2+y^2)^{-2}=-2xy(x^2+y^2)^{-2}
\\ u_{xx} &= -2xy*(x^2+y^2)^{-3}*-2*2x+-2y*(x^2+y^2)^{-2}
     \\&= \frac{-2xy}{(x^2+y^2)^3}*-4x +\frac{-2y}{(x^2+y^2)^2}
     \\&=\frac{8x^2y}{(x^2+y^2)^3}+\frac{-2y}{(x^2+y^2)^2}
\\
\end{aligned} $$
Partial Derivatives with y 
$$\begin{aligned} 
u_y&=1(x^2+y^2)^{-1}+y*(x^2+y^2)^{-2}*-1*2y
\\    &=(x^2+y^2)^{-1}-2y^2(x^2+y^2)^{-2}
\\ u_{yy}&=-1(x^2y^2)^{-2}*2y -2*2y(x^2+y^2)^{-2}-2y^2*-2(x^2+y^2)^{-3}*2y
   \\      &=-2y(x^2+y^2)^{-2}-4y(x^2+y^2)^{-2}+8y^3(x^2+y^2)^{-3}
     \\   &=\frac{-6y}{(x^2+y^2)^2} + 8y^3(x^2+y^2)^{-3}
\end{aligned} $$
From here need to show that $u_{xx}+u_{yy}=0$ and technically say why the other partials are continous right?? This was a test question whith 3 lines of paper by the way
 A: $$
 f(z) = \frac 1z = \frac{1}{x+iy} = \frac{x}{x^2+y^2} + i\frac{-y}{x^2+y^2} 
$$
is holomorphic in $\Bbb C \setminus \{ 0 \}$. It follows that
$$
 -\operatorname{Im} f(z) = \frac{y}{x^2+y^2}
$$
is harmonic in the same domain.
(It is  a direct consequence of the Cauchy-Riemann 
differential equations that real and imaginary part of a holomorphic
function are harmonic.)
A: Let $F(x,y) = \frac{1}2 \log(x^2+y^2)$. Then $f$ is harmonic everywhere away from the origin, and so are its derivatives $\partial_1 F$ and $\partial_2 F$. To prove the first claim, note that 
$$\partial_1 F = \frac{x}{x^2+y^2}$$
$$\partial_{11} F = \frac{y^2-x^2}{(x^2+y^2)^2}$$
and by symmetry
$$\partial_{22} F = \frac{x^2-y^2}{(x^2+y^2)^2} = -\partial_{11}F.$$
A: $$\begin{align*}&u_x=-\frac{2xy}{(x^2+y^2)^2}\;&,\;\;&u_{xx}=-\frac{2y(x^2+y^2)-8x^2y}{(x^2+y^2)^3}=\frac{6x^2y-2y^3}{(x^2+y^2)^3}\\{}\\
&u_y=\frac{x^2+y^2-2y^2}{(x^2+y^2)^2}=\frac{x^2-y^2}{(x^2+y^2)^2}\;&,\;\;&u_{yy}=\frac{-2y(x^2+y^2)-4y(x^2-y^2)}{(x^2+y^2)^3}=\frac{-6x^2y+2y^3}{(x^2+y^2)^3}\end{align*}$$
and now just check that we certainly get $\;\nabla f=u_{xx}+u_{yy}=0\;$ . Observe the function $\;f\;$ is defined everywhere except at the origin, and the partial derivatives above, all of them (first and second order) are continuous everywhere where $\;f\;$ is defined.
A: While I believe that the "right" answers are those already given, I would like to add yet another one, based on polar coordinates. Introduce 
$$
\begin{cases}
x=r\cos \phi\\
y=r\sin \phi
\end{cases}
$$
The given function $f(x, y)=\frac{y}{x^2+y^2}$ is harmonic if and only if 
$$
\left(\partial_r^2 +r^{-1}\partial_r +r^{-2}\partial_\phi^2\right)\left(\sin (\phi) r^{-1}\right)=0, $$ 
which is, of course, true. 
This computation is slightly faster than the Cartesian one because $f$ is separable in polar coordinates. ("Separable" here means that a function is expressed as a product of functions of a single variable).
