Given $f(x,y)=x^2 + 4xy + y^2 - 2x + 2y + 1$
I tried setting $F(x,y,z) = f(x,y) - z$ and found the partial derivatives Fx, Fy, and Fz.
Fx = 2x + 4y -2
Fy = 4x + 2y + 2
Fz = -1
Apparently the gradient is perpendicular to the tangent plane, so it must be parallel to <0,0,1> which the normal vector of the xy plane. So I set Fx = 0 and Fy = 0 and solved for x and y, which turns out to be -1 and 1 respectively. But I don't know where to go from here, or whether what I have done so far is correct at all. Please help.
If $f(x,y)$ has a horizontal tangent plane, that means $$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0.$$ So you calculate each partial derivative, and solve for when both are simultaneously zero. Then plug in those coordinates into $f$ to get the $z$-value.