How to approximate the distance at which the thin disk peaks along a given LOS from Earth for a given galactic coordinate per 1 sq deg?
If you have a ray from an origin point $\mathbf{x}_0$  with unit direction $|\mathbf{v}|=1$,  $\mathbf{x}(s)=[x(s),y(s),z(s)]=\mathbf{x}_0+\mathbf{v}s$ and look at the density times distance $\rho(s)=\rho_0 e^{-R(\mathbf{x})/h_R}e^{-z(\mathbf{x})/h_z}s^2$ the derivative is
$\rho'(s) =\rho_0\left\{ [-e^{-R(\mathbf{x})/h_R}R'(\mathbf{x})/h_R]e^{-z(\mathbf{x})/h_z}s^2 + [-e^{-z(\mathbf{x})/h_z}z'(\mathbf{x})/h_z]e^{-R(\mathbf{x})/h_R}s^2 +[2s]e^{-R(\mathbf{x})/h_R}e^{-z(\mathbf{x})/h_z} \right\}$
$=\rho(s)\left\{[-R'(\mathbf{x})/h_R]+[-z'(\mathbf{x})/h_z]+[2s]/s^2\right\}$.
Now, $R'(s) = (x(s)v_x+y(s)v_y)/R(s)$ and $z'(s)=v_z$ so the criterion for a maximum or minium is $0 = -(x(s)v_x+y(s)v_y)/h_RR(s)-v_z/h_z + 2/s$ or $$\frac{2}{s} = \frac{[x_0v_x+y_0v_y] +[v_x^2 + v_y^2]s}{h_R\sqrt{(x_0+v_xs)^2+(y_0+v_ys)^2}}+\frac{v_z}{h_z}$$
I can't see how to simplify it further.
I think the right approach here is just to do a numeric solution using golden section search rather than do a lot of algebra that anyway ends with a root finding algorithm. Doing that also allows you to use any stellar density model.