Given a family of curves $F(x, y, c) = 0$ for $c$ in a range of real numbers, the envelope $E$ is the curve tangent to every member of the family.
I see that it is defined by the solution of $F(x, y, c) = 0 = \dfrac{\partial F}{\partial c} (x, y, c)$.
How do I prove this ? I know that if $\dfrac{\partial F}{\partial c} (x, y, c) \neq 0$, then we can solve $F(x, y, c) = 0$ at least locally for $c = c(x, y)$ but then I cannot see how $\dfrac{\partial F}{\partial c} (x, y, c) = 0$.
I saw a similar question with an answer that involved some differential geometry and "straightening of fields" that I do not understand. Can I have an answer based on simple analysis that explains
a) How does the solution of the two equations define $E$
b) What is the problem if both $\dfrac{\partial F}{\partial x} = \dfrac{\partial F}{\partial y} = 0$
c) What happens if $F(x, y, c) = f(x, y) + \phi(c)$ ?
First of all, you want the individual curves $F_c(x,y)=F(x,y,c)=0$ for fixed $c$ to be smooth, so you'd better assume that $\dfrac{\partial F}{\partial x}$ and $\dfrac{\partial F}{\partial y}$ are never simultaneously zero. Then you want the implicit function theorem to guarantee you that your two equations will define a smooth curve (locally) parametrized by $c$, so you want the matrix $$\begin{bmatrix} \frac{\partial F}{\partial x} & \frac{\partial F}{\partial y} \\ \frac{\partial^2 F}{\partial x\partial c} & \frac{\partial^2 F}{\partial y\partial c}\end{bmatrix}$$ to be nonsingular at some point $(x_0,y_0,c_0)$.
Then you have to check that the curve your equations define locally parametrically as $(x,y)=\phi(c)$ has the property that it is tangent to the curve $F_c(x,y)=0$ at $\phi(c)$. I'll let you check this yourself by differentiating $F(\phi(c),c) = 0$.
In answer to your third question, you'll get parallel level curves and there will be no envelope. You might want to try doing some examples. Here's one for you: $F(x,y,c)=y+c^2x-c$.