So those are the quantities I want. I have $r_x$, $r_{xx}$, $r_y$, $r_{yy}$
So now $\tan(\theta) = \frac{y}{x}$, take partials wrt $y$
$\sec^2(\theta) \theta_y = \frac{1}{x}$, so $\theta_y = \frac{\cos^2(\theta)}{x} = \frac{x}{r^2}$
And $\theta_{yy} = \frac{-2xy}{r^4}$
Now taking partials wrt $x$ is gonna look different, we get $\sec^2(\theta) \theta_x = -\frac{y}{x^2}$
So $\theta_x = -\frac{y}{r^2}$, and $\theta_{xx} = \frac{2xy}{r^4}$
Conveniently $\theta_{xx} + \theta_{yy} = 0$
And $\theta_x^2 + \theta_y^2 = \frac{1}{r^2}$
So now putting everything together, help me God
I was called now I'm back
$f_{xx} = f_{rr}r_x^2 + 2f_{\theta r}\theta_xr_x + f_{\theta\theta}\theta_x^2 + f_rr_{xx} + f_{\theta}\theta_{xx}$
$f_{yy} = f_{rr}r_y^2 + 2f_{\theta r}\theta_yr_y + f_{\theta\theta}\theta_y^2 + f_rr_{yy} + f_{\theta}\theta_{yy}$
Now we have some convenient facts
$r_{xx}+r_{yy} = \frac{1}{r}$
$\theta_{xx}+\theta_{yy} = 0$
$\theta_x^2 + \theta_y^2 = \frac{1}{r^2}$
$r_x\theta_x + r_y\theta_y = 0$
$f_{xx}+f_{yy} = f_{rr} + \frac{1}{r}f_r + \frac{1}{r^2}f_{\theta\theta}$
So now in polar coordinates, the hyperbolic Laplacian is $(1-r^2)^2(f_{rr} + \frac{1}{r}f_r + \frac{1}{r^2}f_{\theta\theta})$