Let $\mathbb H^2 := \left\{ z = x+iy \in \mathbb C : y > 0\right\}$, equipped with the hyperbolic Riemannian metric $\breve{g} = \frac 1{y^2}\left(dx^2 + dy^2\right)$. It is a classical result that the group $\mathrm{SL}(2,\mathbb R)$ acts on $\mathbb H^2$ isometrically via $$ \left(\begin{array}{cc} a & b \\ c & d \end{array}\right) \cdot z = \frac{az+b}{cz+d} $$ Denote $\theta_A(z) = A\cdot z$ for $A \in \mathrm{SL}(2,\mathbb R)$. Then it is straightforward to calculate $$ \frac{d}{dz}\theta_A(z) = \frac 1{(cz+d)^2}. $$ Problem: Find a complex-analytic proof using $\frac{d}{dz}\theta_A(z)$ to show $\theta_A : \mathbb H^2 \to \mathbb H^2$ is an isometry.
I have the following argument in real Cartesian coordinates: $\theta_A$ becomes $\theta_A(x,y) = \big( u(x,y),\: v(x,y)\big)$, where $$ u(x,y) = \frac{(ax+b)(cx+d)+acy^2}{(cx+d)^2+c^2y^2},\qquad v(x,y) = \frac y{(cx+d)^2+c^2y^2} $$ Since $\theta_A : \mathbb H^2 \to \mathbb H^2$ is a holomorphic function, $u$ and $v$ satisfy the Cauchy-Riemann equations, so the differential of the map $\theta_A : \mathbb H^2 \to \mathbb H^2$ (considering $\mathbb H^2 \subset \mathbb R^2$) satisfies: \begin{align*} d(\theta_A)_z &= \left(\begin{array}{cc} u_x & u_y \\ v_x & v_y \end{array}\right) = \left(\begin{array}{cc} v_y & -v_x \\ v_x & v_y \end{array}\right) \\ &= \frac 1{\left((cx+d)^2+c^2y^2\right)^2}\left(\begin{array}{cc} (cx+d)^2-c^2y^2 & 2cy(cx+d) \\ -2cy(cx+d) & (cx+d)^2 - c^2 y^2 \end{array}\right) \end{align*} Therefore if we let $\left\{ \partial_x, \partial_y\right\}$ denote the standard coordinate frame induced by the identity coordinates $\mathbb H^2 \hookrightarrow \mathbb R^2$, it's a starightforward calculation to show $$ \theta_A^*\breve g(\partial_x, \partial y) = 0, \qquad \theta_A^* \breve g(\partial_x, \partial_x) = \theta_A^* \breve g(\partial_y, \partial_y) = \frac 1{y^2}, $$ whence it follows that $\theta_A^* \breve{g} = \breve{g}$, so $\theta_A$ is a Riemannian isometry.
However: I'm trying to develop comfort and familiarity with complex one-forms and Riemannian geometry of complex surfaces, so I would like to know what the details would be in order to use $\theta_A'(z) = (cz+d)^{-2}$ to show $\theta_A$ is an isometry, without falling back to Cartesian coordinates. I'd also like to avoid using geodesics if possible, in favor of using complex differential forms and complex coordinate frames (i.e. $\left\{ dz, d\overline z\right\}$ and $\left\{\partial/\partial z, \partial/\partial\overline z\right\}$ respectively). Any insights on what this flavor of argument would look like?
EDIT: See comments for a solution.