I want to show that the following Riemannian manifold with given metric $g$ the following vector fields are orthonormal at every point $p$ of the manifold.
Let $\mathcal{H}^3=\lbrace (x,y,z) \in \mathbb{R}^3 \mid z>0\rbrace$ be equipped with the Riemannian metric: \begin{equation*} g=\frac{dx^2+dy^2+dz^2}{z^2} \end{equation*} And the vector fields $A,B,C$ defined as: \begin{equation*} A=z\partial_x, \quad B=z\partial_y, \quad C=z\partial_z \end{equation*}
Trying to plug the vector spaces into the definition of $g$ gives:
\begin{equation} g(A,B)=\frac{dx(z\partial_x)dx(z\partial_y)+dy(z\partial_x)dy(z\partial_y)+dz(z\partial_x)dz(z\partial_y)}{z^2} \end{equation}
But since all the terms in the numerator vanish this expression is zero.
The way in which $g$ is written is maybe the cause for the confusion. Try writing it like this:
$$g(X,Y)=\frac{dx(X)dx(Y)+dy(X)dy(Y)+dz(X)dz(Y)}{z^2} $$
Now it should be clear that $dx(\partial_x)=dz(\partial_z)=dy(\partial_y)=1$ and $dy(\partial_x)=0$ which also holds for all other combinations (i.e. $dy(\partial_z)$ etc.).
Can you figure it out from here on or do you need more help?