The problem statement is
Show that every local isometry is conformal.
I have seen this question posed several times, in particular here
This is Problem 6.3.1 in Pressley's Elementary Differential Geometry book.
I used the fact that,
a local diffeomorphism $f: S_1 \rightarrow S_2$ is a local isometry if and only if, for any surface patch $\sigma_1$ of $S_1$, the patches $\sigma_1$ and $f \circ \sigma_1$ of $S_1$ and $S_2$, respectively have the same first fundamental form.
a local diffeomorphism $f: S_1 \rightarrow S_2$ is conformal if and only if, for any surface patch $\sigma$ of $S_1$, the first fundamental forms of the patches $\sigma$ of $S_1$and $f \circ \sigma$ of $S_2$ are proportional.
This totally makes sense. The constant of proportionality is 1. But then the author goes on to say that
In particular, a surface patch $\sigma(u,v)$ is conformal if and only if its first fundamental form is $\lambda(du^2+dv^2)$ for some smooth function $\lambda(u,v)$
This is my source of confusion. From the last statement, I am reading that the terms of the first fundamental form are $F_1=F_2=0$ and $E_1=E_2=G_1=G_2$, but the condition of an isometry is $E_1=E_2, F_1=F_2$, and $G_1=G_2$. The isometry condition is less restrictive. So how does having the condition of an isometry imply we are also satisfying the condition of conformal? How can I say that $E=G$ and $F= 0$?