Prove that the coordinate lines of a surface patch are lines of curvature if and only if $f=F=0$.
Lines of curvature being defined as follows: A unit-speed curve $\gamma:I\rightarrow S$ in an oriented regular surface S is called a line of curvature if $\gamma'(t)$ is a principle direction for all $t∈ I$
I started with the equation for Gaussian curvature $K=\frac{eg-f^2}{EG-F^2}$, and with $f=F=0$, I have $K=\frac{eg}{EG}=\frac{e}{E}*\frac{g}{G}$
And Mean curvature $H=\frac{eG-2fF+gE}{2(EG-F^2)}=\frac{eg+gE}{2EG}=\frac{e}{E}+\frac{g}{G}$
Principle curvatures $k_1$ and $k_2$ as $k_1 +k_2=2H, k_1k_2=K$.
I'm having trouble understanding how to incorporate $\gamma$ as a principle direction and beginning the proof. Any direction of help would be great, thanks.
Note that for a patch $x\colon U\subset \mathbb{R}^2\to\mathbb{R}^3$, the vectors $x_u$ and $x_v$ are the tangent vectors of the $u$- and $v$-lines. A $u$-line is parametrized by $\gamma(t)= x(t,v_0)$ with $v_0$ constant; $v$-lines are parametrized by $\gamma(t)=x(u_0,t)$ with constant $u_0$.
What do you now have to show? Well, you need to show that, at every point, $x_u$ and $x_v$ are eigenvectors of the shape operator. For this, you have to show that the matrix of the shape operator (w.r.t. the basis $x_u$, $x_v$) is diagonal.
Can you take it from here?