The surface in question is
$$S=\{(x,y,z)\in\mathbb{R}^3:x^2-yz^{-2}=0\}$$
which can be rewritten as
$$S=\{(x,y,z)\in\mathbb{R}^3:z\neq 0\ \text{and} \ x^2z^2-y=0\}$$
Put $h(x,y,z)=x^2z^2-y\implies \nabla h(x,y,z)=(2xz^2,1,2x^2z)$. Now, a vector is parallel to the $xy$ plane if and only if it is perpendicular to $(0,0,1)$, and
$$\nabla h(x,y,z)\perp (0,0,1)\iff (2xz^2,1,2x^2z)\cdot(0,0,1)=0\iff 2x^2z=0.$$
Since $z\neq0$ for $(x,y,z)\in S$, it follows that $x=0$, and therefore $y=0$. Hence, any point of the form $(0,0,z)$ with $z\neq 0$ is a solution.
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 $(2,-1,4), (-2,1,-4)$.
Best Answer
Let $h(x,y,z)=2x^2+y^4+1-z^2$ and let $S^\star=\{(x,y,z)\in\Bbb R^3\mid h(x,y,z)=0\}$; then $S=\{(x,y,z)\in S^\star\mid z>0\}$. You have $\nabla h(x,y,z)=(4x,4y^4,-2z)$ and you want to find a point of $S$ such that $\nabla h(x,y,z)$ is a multiple of $(1,-1,-1)$. So, you solve the system:$$\left\{\begin{array}{l}4x=\lambda\\4y^3=-\lambda\\-2z=-\lambda\\h(x,y,z)=0.\end{array}\right.$$It is not hard to see that the only solution is $x=1$, $y=-1$, $z=2$, and $\lambda=4$. So, the plane that you're after is the plane passing through $(1,-1,2)$ which is parallel to the plane $x-y-z=0$, which is the plane $x-y-z=0$ itself.