What points on surface is the tangent plane parallel to $xy$- plane? At what pts. on the surface $z = x^{2}y + y^{2}x + 3x$ is the tangent plane parallel to the $xy$-plane? 
So first I define a function $F(x, y, z) =  x^{2}y + y^{2}x + 3x - z$ which has gradient $grad F = (2xy + y^2 + 3, x^{2} + 2yx, -1)$. So the equation of our tangent plane is: $(2xy + y^{2} + 3)(x - x_{0}) + (x^{2} + 2yx)(y - y_{0}) - (z - z_{0}) = 0$.
So we get as a normal line:
$r(t) = (x_{0} + (2x_{0}y_{0} + y_{0}^{2} + 3)t, y_{0} + (x_{0}^{2} +2y_{0}x_{0})t, z_{0} - t)$. 
Now I know two planes are parallel if their normal lines are parallel, but I'm not quite sure how to complete the problem.

If the tangent plane to the surface defined by $f(x,y,z) = 0$ with $f(x,y,z) = x^2y+y^2x+3x-z$ at the point $P = (x,y,z)$ is parallel to the $xy$ plane, then its normal line must be parallel to the vector $\vec{a} = <0,0,1>$. But $\vec{n} = \nabla{f}= <2xy+y^2+3, x^2+2xy, -1>$. Thus we must have: $2xy+y^2+3 = 0 = x^2+2xy$. Observe that $x \neq 0$ since $y^2+3 > 0$. Thus: $x(x+2y) = 0 \implies x = -2y\implies -4y^2+y^2+3 = 0\implies y^2 = 1 \implies y = \pm 1\implies x = \mp 2$. Thus the points are $(2,-1,4), (-2,1,-4)$.
