Just as the envelope of a $1$-parameter family of curves in the plane is defined, now define the envelope of a $2$-parameter family of surfaces in $\Bbb R^3$. The envelope should be a surface so that each member of the family is tangent to the surface at some point.
Analytically, if $U\subset\Bbb R^2$ is an open subset and we're given a $C^2$ function $f\colon \Bbb R^3\times U\to\Bbb R$ with $\nabla_{\mathbf x} f(\mathbf x,u,v)\ne \mathbf 0$ at each point of the level surface $S_{(u,v)} = \{\mathbf x: f(\mathbf x,u,v)=0\}$, then you can check that the envelope comes from solving the system
$$f(\mathbf x,u,v)= \frac{\partial f}{\partial u}(\mathbf x,u,v) = \frac{\partial f}{\partial v}(\mathbf x,u,v) = 0.$$
There is a nondegeneracy hypothesis$^*$ that allows you to deduce from the Implicit Function Theorem that you'll locally have a smooth surface defined by these equations.
If on some open subset of $U$ we have $\mathbf x = \Phi(u,v)$ given by these equations, then we claim that this equation defines a parametric surface $S$. We now check that at the point $\Phi(u,v)$ it is tangent to the surface $S_{(u,v)}$. Set $g(u,v) = f(\Phi(u,v),u,v)$ and note that since $g$ is identically $0$, we have
$$\frac{\partial g}{\partial u} = \nabla_{\mathbf x} f\cdot\frac{\partial\Phi}{\partial u} + \frac{\partial f}{\partial u} = 0, \tag{$\dagger$}$$
and similarly for the $v$ partial derivative. Thus, we conclude that
$$\nabla_{\mathbf x} f\cdot\frac{\partial\Phi}{\partial u} = \nabla_{\mathbf x} f\cdot\frac{\partial\Phi}{\partial v} = 0.$$
Since $\nabla_{\mathbf x}f(\mathbf x,u,v)$ is the normal to the level surface $S_{(u,v)}$ and since $\frac{\partial\Phi}{\partial u}$ and $\frac{\partial\Phi}{\partial v}$ span the tangent plane of $S$, we are done.
$^*$For those who are interested, here's the hypothesis. Now we write $\mathbf x = (x,y,z)$. Assume that the matrix
$$\left[\begin{matrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \\
\frac{\partial^2 f}{\partial x\partial u} & \frac{\partial^2 f}{\partial y\partial u} & \frac{\partial^2 f}{\partial z\partial u} \\
\frac{\partial^2 f}{\partial x\partial v} & \frac{\partial^2 f}{\partial y\partial v} & \frac{\partial^2 f}{\partial z\partial v} \end{matrix}\right]$$
is nonsingular. This will give you a local representation of the solution of ($\dagger$) as a $C^1$ function of $(u,v)$.