Find the equations of the curves that cut the angle between the coordinate lines of the surface in half.
The surface being: $$x = u \cos v \\ y=u \sin v \\ z=u ;\\ (u>0)$$
So, we have the parameterization immediately: $(u>0)$ $$(u,v)\to(u\cos v, u \sin v, u) $$
The coordinate lines being $\ \ v=v_0; \ u =u_0; \ \ \ \ \ \ v_0,u_0=const.$
The parametrisation sets an orthogonal coordinate system in the surface, so, we look for curves for wich their tangent vector is of the form $T=at_u+bt_v$ and having all along them $a/b=1$, with $t_u$ and $t_v$ unitary vectors tangent to the curves with $v$ constant and $u$ constant respectively.
$T_u=(\dfrac{\partial x}{\partial u},\dfrac{\partial y}{\partial u},\dfrac{\partial z}{\partial u})$; $t_u=\dfrac{T_u}{\vert T_u\vert}=\dfrac{1}{\sqrt{2}}(\cos v, \sin v,1)$
$T_v=(\dfrac{\partial x}{\partial v},\dfrac{\partial y}{\partial v},\dfrac{\partial z}{\partial v})$; $t_v=\dfrac{T_v}{\vert T_v\vert}=\dfrac{1}{u}(-u\sin v, u\cos v,0)$
$\left(t_u·t_v=0\right.$ confirming the orthogonality.$\left.\right)$
The vector tangent to any curve in the surface is:
$T(u(t),v(t))=\dfrac{dr}{dt}=T_u\dfrac{du}{dt}+T_v\dfrac{dv}{dt}=\sqrt{2}u't_u+uv't_v$ ($u'=du/dt$ and $v'=dv/dt$)
And imposing the condition to cut the (right) angle in two:
$\dfrac{uv'}{\sqrt{2}u'}=1$ or $\dfrac{v'}{u'}=\dfrac{\sqrt{2}}{u}$
$\dfrac{dv}{du}=\dfrac{\sqrt{2}}{u}\implies u=Ae^{v/\sqrt{2}}$
The curves are thus,
$r=(Ae^{v/\sqrt{2}}\cos v,Ae^{v/\sqrt{2}}\sin v,Ae^{v/\sqrt{2}})$