Show that $z = \ln (x^2+y^2) +2\tan^{-1}(y/x)$ satisfies the laplaces's equation. 
I can get to the first partial derivative of $\partial z / \partial x$ of course, but after that I get this, which I don't know how to differentiate further... I am talking about c part of the question. Attaching my working...


Differentiating once:
$$\frac{\partial z}{\partial x}\left( \ln(x^2+y^2)+2\tan^{-1}\left( \frac{y}{x}\right )\right )=\frac{2x}{x^2+y^2}+\frac{-2\frac{y}{x^2}}{\frac{x^2+y^2}{x^2}}=\frac{2x}{x^2+y^2}-\frac{2y}{x^2+y^2}=\frac{2x-2y}{x^2+y^2}$$
Differentiating again:
$$\frac{\partial z}{\partial x}\left ( \frac{\partial z}{\partial{x}}\right )=\frac{\partial}{\partial x}\left( \frac{2x-2y}{x^2+y^2}\right )=\frac{2(x^2+y^2)-2x(2x-2y)}{(x^2+y^2)^2} \\ =\boxed{\frac{-2x^2+2y^2+2xy}{(x^2+y^2)^2}}$$

Now let us show $\frac{\partial^2 z}{\partial y^2}$:
$$\frac{\partial}{\partial y}\left( \ln(x^2+y^2)+2\tan^{-1}\left( \frac{y}{x}\right )\right )=\frac{2y}{x^2+y^2}+\frac{\frac{1}{x}}{1+(\frac{y}{x})^2}=\frac{2y+x}{x^2+y^2}$$
$$\frac{\partial }{\partial y}\left( \frac{2y+x}{x^2+y^2}\right )=\frac{2(x^2+y^2)-(2y)(2y+x)}{(x+y)^2}=\boxed{\frac{2x^2-2y^2-2xy}{(x^2+y^2)^2}}$$

Therefore,
$$\frac{\partial^2 z}{\partial x^2}+\frac{\partial ^2 z}{\partial y^2}=\frac{-2x^2+2y^2+2xy}{(x^2+y^2)^2}+\frac{2x^2-2y^2-2xy}{(x^2+y^2)^2}=0 \checkmark$$
