How to solve $\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + f = 0$ $\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + f = 0$. (1)
By putting 
$x = r\cos\theta, y = r \sin \theta$, 
$g(r,\theta) = f(x(r,\theta), y(r,\theta))$, 
we have
$\frac{\partial^2 f}{\partial x^2} = \cos^2\theta\frac{\partial^2g}{\partial r^2} - \frac{\sin2\theta}{r}\frac{\partial^2 g}{\partial r \partial \theta} +\frac{\sin^2\theta}{r^2} \frac{\partial^2 g}{\partial\theta^2}$ 
$\frac{\partial^2 f}{\partial y^2} = \sin^2\theta\frac{\partial^2g}{\partial r^2} + \frac{\sin2\theta}{r}\frac{\partial^2 g}{\partial r \partial \theta} +\frac{\cos^2\theta}{r^2} \frac{\partial^2 g}{\partial\theta^2}$ 
(because $\frac{\partial}{\partial x} = \cos\theta\frac{\partial}{\partial r} - \frac{\sin\theta}{r}\frac{\partial}{\partial \theta}$, $\frac{\partial}{\partial y} = \sin\theta\frac{\partial}{\partial r} + \frac{\cos\theta}{r}\frac{\partial}{\partial y}$).
By substituting this for (1) he have
$\frac{\partial^2 g}{\partial r^2} + \frac{1}{r^2}\frac{\partial^2 g}{\partial \theta^2} + g = 0$.
By substituting $g(r,\theta) = R(r)\sin\theta$, we have
$\frac{d^2 R}{dr} + (1-\frac{1}{r^2})R = 0.$
But my print says that this should be 
$\frac{d^2 R}{dr} +\frac{1}{r}\frac{dR}{dr}+ (1-\frac{1}{r^2})R = 0.$
I could not find a mistake. Please advise me. 
Sorry for my bad English.
 A: You wrote :

$\frac{\partial^2 f}{\partial x^2} = \cos^2\theta\frac{\partial^2g}{\partial r^2} - \frac{\sin2\theta}{r}\frac{\partial^2 g}{\partial r \partial \theta} +\frac{\sin^2\theta}{r^2} \frac{\partial^2 g}{\partial\theta^2}$
$\frac{\partial^2 f}{\partial y^2} = \sin^2\theta\frac{\partial^2g}{\partial r^2} + \frac{\sin2\theta}{r}\frac{\partial^2 g}{\partial r \partial \theta} +\frac{\cos^2\theta}{r^2} \frac{\partial^2 g}{\partial\theta^2}$
(because $\frac{\partial}{\partial x} = \cos\theta\frac{\partial}{\partial r} - \frac{\sin\theta}{r}\frac{\partial}{\partial \theta}$, $\frac{\partial}{\partial y} = \sin\theta\frac{\partial}{\partial r} + \frac{\cos\theta}{r}\frac{\partial}{\partial y}$).

Some terms are missing and there is a typo in the third equation. The correct writing is :
$\frac{\partial^2 f}{\partial x^2} = \cos^2\theta\frac{\partial^2g}{\partial r^2} - \frac{\sin2\theta}{r}\frac{\partial^2 g}{\partial r \partial \theta} +\frac{\sin^2\theta}{r^2} \frac{\partial^2 g}{\partial\theta^2}+
\frac{\sin^2\theta}{r} \frac{\partial g}{\partial r}+\frac{\sin 2\theta}{r^2} \frac{\partial g}{\partial\theta}$
$\frac{\partial^2 f}{\partial y^2} = \sin^2\theta\frac{\partial^2g}{\partial r^2} + \frac{\sin2\theta}{r}\frac{\partial^2 g}{\partial r \partial \theta} +\frac{\cos^2\theta}{r^2} \frac{\partial g}{\partial r}
+\frac{\cos^2\theta}{r} \frac{\partial g}{\partial r}-\frac{\sin 2\theta}{r^2} \frac{\partial g}{\partial\theta}$
(because $\frac{\partial}{\partial x} = \cos\theta\frac{\partial}{\partial r} - \frac{\sin\theta}{r}\frac{\partial}{\partial \theta}$, $\frac{\partial}{\partial y} = \sin\theta\frac{\partial}{\partial r} + \frac{\cos\theta}{r}\frac{\partial}{\partial \theta}$).
Since your calculus is not detailed, one cannot say where the terms were lost.
Then adding the two first equation and after simplification :
$$\boxed{\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}\:=\:\frac{\partial^2 g}{\partial r^2}+\frac{1}{r}\frac{\partial g}{\partial r}+\frac{1}{r^2}\frac{\partial^2 g}{\partial \theta^2}}$$
This is the correct Laplacian in polar coordinates which can be found in many books. http://mathworld.wolfram.com/Laplacian.html
Application to the next PDE :
$$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + f = 0$$
$$\frac{\partial^2 g}{\partial r^2}+\frac{1}{r}\frac{\partial g}{\partial r}+\frac{1}{r^2}\frac{\partial^2 g}{\partial \theta^2}+g=0$$
I suppose that you are looking for particular solutions (not general solution) on the form
$$g(r,\theta) = R(r)\sin\theta$$
$$\sin\theta \frac{d^2R}{dR^2}+\sin\theta \frac{1}{r}\frac{dR}{dr}+\frac{1}{r^2}(-\sin\theta )R+\sin\theta R=0$$
$$\frac{d^2R}{dR^2}+ \frac{1}{r}\frac{dR}{dr}+\left(1-\frac{1}{r^2}\right)R=0$$
