Let $\Bbb{H}:=\{(x,y)\in \Bbb{R}^2: y>0\}$ be the hyperbolic halfplane induced with the hyperbolic riemannian metric $$(g_\Bbb{H})_{(x,y)}=\frac{1}{y^2}\left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) $$ I consider the euclidean plane $\Bbb{R}^2$ with the euclidean metric $g=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) $. I want to show that the identity map $$\operatorname{id}:(\Bbb{H}, g_\Bbb{H})\rightarrow (\Bbb{R},g)$$ is conformal.
By definition I need to find $\lambda: \Bbb{H}\rightarrow \Bbb{R}_{>0}$ such that $$g_{(x,y)}(D\operatorname{id}|_{(x,y)} X,D\operatorname{id}|_{(x,y)} Y)=\lambda^2(x,y)(g_\Bbb{H})_{(x,y)}(X,Y)$$ for all $(x,y)\in \Bbb{H}$ and for all $X,Y\in T_{(x,y)}\Bbb{H}$.
Proof Let $(x,y)\in \Bbb{H}$, i.e. $y>0$. Then $$\begin{align}(g_\Bbb{H})_{(x,y)}(X,Y)=\frac{1}{y^2}(X,Y)\left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right)\left( \begin{matrix} X \\ Y \end{matrix} \right)\end{align}$$ On the other hand $$\begin{align}g_{(x,y)}(D\operatorname{id}|_{(x,y)} X,D\operatorname{id}|_{(x,y)} Y)&=g_{(x,y)}(X,Y)\\&=(X,Y)\left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right)\left( \begin{matrix} X \\ Y \end{matrix} \right)\end{align}$$ So we see that we can define $\lambda: \Bbb{H}\rightarrow \Bbb{R}_{>0};~~(x,y)\mapsto \frac{1}{y}$ then $\lambda^2(x,y)=\frac{1}{y^2}$ and we get the equality we wanted.
Is this mathematically correct?