The surface is flat if its Guassian curvature vanishes everywhere. The surface is minimal if ist mean curvature vanishes everywhere. Therefore, for a function f(x,y), its corresponding surface's Gaussian curvature and mean curvature are both 0.
My question is how to show its second fundamental form Ldu^2 + 2Mdudv + Ndv^2 = 0?
HINT:
If you meant $r(u,v)$ then due to orthogonality between normal N and tangent vectors we have
$$ N \cdot r_1=0 $$
Differentiate w.r.t. $u$ we have by Chain Rule $ \; N\cdot r_{11} + N_1 \cdot r_{1} =0 $
and since $ (N_1, r_1) $ are orthogonal their dot product vanishes leaving $ (r_{11}\cdot N) =L=0$. Differentiation w.r.t. $v$ leaves $M=0$.
Similarly $ N \cdot r_2=0 $, again differentiate w.r.t $v$ in $ N\cdot r_{22} +N_2\cdot r_{2} $ second term vanishes leaving the first term $ (M= r_{22}\cdot N=0 )$ so that
$$ L du^2+ 2 M du dv + N dv^2 =0 ;\ $$
Moreover in the plane i.e., principal scalar curvatures are all zero
$$ \{ k_1k_2=0, k_1+k_2=0\} \to \{k_1=0,k_2=0\} .$$