How is the differential equation $\frac{∂^2f}{∂x^2} + \frac{∂^2f}{∂y^2} = 0$ transformed if you switch to polar coordinates, x = r cos θ, y = r sin θ?
I have done everything right except $\frac{∂^2f}{∂\theta^2}$ (I hope you can see it clearly):
However this is how they did $\frac{∂^2f}{∂\theta^2}$ in my book:
The thing I don't understand is where they get $-r\frac{∂z}{∂r}$ from in their solution for $\frac{∂^2f}{∂\theta^2}$? Or is my answer correct? Did I answer what I was asked?