In Example 3 of section 5.10 of Do Carmo's book "Differential Geometry of Curves and Surfaces", we have that the curvature of the (Lobachevsky) hyperbolic plane is
$$K=\frac{1}{2\sqrt{EG}}\bigg\{\left(\frac{E_v}{\sqrt{EG}}\right)_v+\left(\frac{G_u}{\sqrt{EG}}\right)_u\bigg\}$$
which is shown to be equal to -1, since $E=G=1/v^2$ and $F=0$. So, we have $\frac{1}{2\sqrt{EG}}=\frac{1}{2E}=\frac{v^2}{2}$, $G_u=0$ and $E_v=-\frac{2}{v^3}$.
The crucial result to prove that when $F=0$, the parametrization is orthogonal, so we can compute the Gaussian curvature as $$K=\frac{1}{2\sqrt{EG}}\bigg\{\frac{\partial}{\partial u}\frac{G_v}{\sqrt{EG}}+\frac{\partial}{\partial v}\frac{E_v}{\sqrt{EG}}\bigg\}.$$
Could someone please tell me why that is the case?
You just have to slog through the computations using the Christoffel symbols (which simplify somewhat when $F=0$) and the Gauss equation. This is a standard (not particularly fun) exercise. See section 4-3 of DoCarmo.
(If you know about differential forms and the method of moving frames, then the computation becomes quite simple.)