The difference between "z= f(x, y)" and "F(x, y, z)= 0" is that you are taking F(x, y, z)= f(x, y)- z. Taking the gradient \nabla F= <f_x, f_y, -1>[.
It is correct that F(1, -1, 4\sqrt{2})= \frac{1}{4}+ \frac{1}{4}+ \frac{32}{64}= 1 but that does NOT say "z= 1". You are told that at this point z= 4\sqrt{2}. The fact that F(1, -1, 4\sqrt{2})= 1 just verifies that the given point is on the given surface.
The tangent plane is z- z_0= f_x(x_0, y_0)(x- x_0)+ f_y(x_0, y_0)(y- y_0) only applies if you are given z= f(x, y) which is NOT the case here. You are given, rather, F(x, y, z)= x^2/4+ y^2/4+ z^2/64= 1. In order to convert that to the "z= f(x,y)" form you need to solve for z: z^2/64= 1- x^2/4- y^2/4, z/8= \sqrt{1- x^2/4- y^2/4}, z= 8\sqrt{1- x^2/4- y^2/4}.
Much simpler is to use the fact that if F(x, y, z)= 1 (or any constant) then \nabla F is perpendicular to the surface at any point. Since for any point (x, y, z) on the tangent plane to the surface at (x_0, y_0, z_0), the vector (x- x_0)\vec{i}+ (y- y_0)\vec{j}+ (z- z_0)\vec{k} is in the tangent plane, we must have \nabla F\cdot [(x- x_0)\vec{i}+ (y- y_0)\vec{j}+ (z- z_0)\vec{k}]= 0 which is the same as F_x(x_0, y_0, z_0)(x- x_0)+ F_y(x_0, y_0, z_0)(y- y_0)+ F_z(x_0, y_0, z_0)(z- z_0)= 0 as the equation of the tangent plane.