I would like to prove that for some regular surface in $R^3$ parametrized by $x(u,v)$ that has coefficients of first fundamental form $E=G$, $F=0$ that $A=x_{uu}+x_{vv}$ is parallel to the normal vector $N$. I attempted to prove this by first taking $N=x_{u}\wedge x_{v} $ and then taking the cross product $A \wedge N$ and trying to prove it is zero. Expanding and using the given values $E,G,F$ I was able to reduce it to:
$A \wedge N = (x_u - x_v)(X_{vv}\bullet X_v + X_{uu}\bullet X_v) $
I am not sure how I should proceed to prove the second parenthesis to be 0, assuming I haven't made a mistake so far. Any help is appreciated.
I can't get the expression that you have. Are you using the expression for $A\wedge (B\wedge C)$ as a linear combination of $B$ and $C$? That's probably the most productive approach. Then I get $$A\wedge N = \big((X_{uu}+X_{vv})\cdot X_v\big)X_u - \big((X_{uu}+X_{vv})\cdot X_u\big) X_v,$$ so it's close to what you have. At any rate, you should be able to differentiate the equations $E=G$ and $F=0$ to deduce that those coefficients are both $0$.