Let $f : U\subset \mathbb{R}^3\to \mathbb{R}$ be given by $f(x,y,z)=\cos x\sin y e^z$ then we have that your surface is indeed a level set $M = f^{-1}(0)$. Then it's easy: remember that the gradient of a function is orthogonal to the level sets. Using this we have that
$$\nabla f(x,y,z)=(-\sin x\sin ye^z, \cos x\cos y e^z, \cos x\sin y e^{z})$$
So that at $(\pi/2,1,0)$ we have $\nabla f(\pi/2,1,0)=(-\sin 1,0,0)$, so that the normal is a multiple of the vector $e_1$, and hence since the magnitude of the normal vector doesn't matter, we can pick the normal vector to be $e_1$. Of course then, the tangent plane is just the $yz$ plane.
After applying a rigid motion to $S$ and $P$, we may as well assume $(x_0,y_0,z_0)=(0,0,0)$ and $P$ is the $xy$-plane. Let $U$ be a small neighborhood of the origin in $S$; by taking $U$ to be small enough, we may assume that $U$ is the image of an open disk in $\mathbb R^2$ under a local parametrization, and thus $U$ is homeomorphic to a disk. The hypothesis implies that the origin is the only point of $U$ for which the $z$-coordinate is zero, so $U\smallsetminus\{(0,0,0)\}$ is contained in the set where $z\ne 0 $. Since a disk minus a point is connected, it follows that either $z>0$ on all of $U\smallsetminus\{(0,0,0)\}$ or $z<0$ on all of $U\smallsetminus\{(0,0,0)\}$. After reflecting in the $xy$-plane, we may assume it is $z>0$.
Now suppose $v$ is a tangent vector to $S$ at the origin. Then there is a smooth curve $c:(-\varepsilon,\varepsilon)\to U$ satisfying $c(0) = (0,0,0)$ and $c'(0) = v$. If we write $c(t) = (x(t),y(t),z(t))$, we see that $z(t)$ attains a minimum at $t=0$, and therefore $z'(0)=0$. This means that every tangent vector at the origin has zero $z$-coordinate, so $T_{(0,0,0)}S$ is contained in the $xy$-plane (i.e., in $P$). Since both $T_{(0,0,0)}S$ and $P$ are two-dimensional, they must be equal.
Best Answer
Let $F(x,y,z)=xyz-a^3$.Then we can have $$F_x=yz,F_y=xz,F_z=xy.$$ so the normal vector at $(x_0,y_0,z_0)$ is $(y_0z_0,x_0z_0,y_0z_0)=a^3(\dfrac{1}{x_0},\dfrac{1}{y_0},\dfrac{1}{z_0})$.
The tangent plane is $F_x(x-x_0)+F_y(y-y_0)+F_z(z-z_0)=0 \Rightarrow \dfrac{x}{x_0}+\dfrac{y}{y_0}+\dfrac{z}{z_0}=3$.