Show that the equation of the tangent plane at $p=(x_0,y_0,z_0)$ of a regular surface is given by $f(x,y,z)=0$ where $0$ is a regular value of $f$, is
$$f_x(p)(x-x_0)+f_y(p)(y-y_0)+f_z(p)(z-z_0)$$
Chapter 2.4 Exercise 1 in Do Carmo (Tangent Plane)
differential-geometrysurfaces
Related Solutions
If a surface is defined by a regular function $z=g(x,y)$, the equation of the tangent plane at $(x_{0},y_{0},z_{0})$ in terms of the partial derivatives of $g$ is
$$z=z_{0}+\left. \frac{\partial g}{\partial x}\right\vert _{(x_{0},y_{0})}(x-x_{0})+\left. \frac{\partial g}{\partial y}\right\vert _{(x_{0},y_{0})}(y-y_{0}).$$
If the surface is defined implicitly by $f(x,y,z)=0$, then $z=g(x,y)=f^{-1}(0)$ (i.e. $f(x,y,g(x,y))\equiv 0$). Since
$$\frac{\partial g}{\partial x}=-\frac{\partial f}{\partial x}/\frac{\partial f}{\partial z}$$
and
$$\frac{\partial g}{\partial y}=-\frac{\partial f}{\partial y}/\frac{\partial f}{\partial z},$$
the equation of the tangent plane at $(x_{0},y_{0},z_{0})$ is given by
$$z=z_{0}-\left(\frac{\partial f}{\partial x}/\frac{\partial f}{\partial z}\right)_{(x_{0},y_{0},z_{0})}(x-x_{0})-\left(\frac{\partial f}{\partial y}/\frac{\partial f}{\partial z}\right)_{(x_{0},y_{0},z_{0})}(y-y_{0})$$
or
$$(x-x_{0})\left(\frac{\partial f}{\partial x}\right)_{(x_{0},y_{0},z_{0})}+(y-y_{0})\left(\frac{\partial f}{ \partial y}\right)_{(x_{0},y_{0},z_{0})}+(z-z_{0})\left(\frac{\partial f}{\partial z}\right)_{(x_{0},y_{0},z_{0})}=0.$$
In compact notation, $\mathbf{x}=\left( x,y,z\right) ,\mathbf{x}_{0}=\left( x_0,y_0,z_0\right) $, we get
$$\left( \mathbf{x}-\mathbf{x}_{0}\right) \cdot \mathbf{\nabla }f(\mathbf{x}_{0})=0.$$
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.
Best Answer
Let $S=\{(x,y,z)\in\mathbb{R}^3; f(x,y,z)=0\}$, and $p_0\in S$. Now, there exist a curve $\alpha:(-\epsilon,\epsilon)\to S$, such that $\alpha(0)=p_0$ and $\alpha'(0)=w\in T_{p_0}S$, with $w=p-p_0$. Note that $$(f\circ\alpha)(t)=f(x(t),y(t),z(t))=0\implies df_{p_0}(w)=0$$ So the inner product of an element of $S$ and any element of $T_{p_0}S$ is $0$, i.e., $$\langle(f_{x}(p_0),f_{y}(p_0),f_{z}(p_0)),w\rangle=\langle(f_{x}(p_0),f_{y}(p_0),f_{z}(p_0)),p-p_0\rangle=0$$
Where $p=(x,y,z)\in T_{p_0}S$ and $p_0=(x_0,y_0,z_0)\in S$.