I am told very specifically that a cone has the following definition:
$$C_1 = \{(x,y,z) \in \mathbb{R^3} | z^2 = x^2 + y^2 , z > 0\}$$
Please note the $z>0$ part. I want to map it to a cylinder of the form:
$$C_2 = \{(x,y,z) \in \mathbb{R^3} | x^2 + y^2 = 1\}$$
If $f:C_1 \rightarrow C_2$ is smooth, inverse exist and is smooth, then we are in business! I know that in topology we would say that the point in the cone would make these two not topologically equivalent, but the definition here is excluding that point. So would the following map work:
$$f(x,y,z) = (\frac{x}{z}, \frac{y}{z}, z)$$
Clearly, $(x/z)^2 + (y/z)^2 = 1$ as required. furthermore, the function is differentiable in $z$ as long as $z > 0$. The inverse also exists as:
$$f^{-1}(u,v,w) = (uw,vw,w)$$
Which is clearly smooth wrt all variables. Hence, am I correct in concluding that these two are diffeomorphic?
Your map $f$ is quite clearly not surjective as $f(x, y, z)$ has positive $z$-coordinates (so, e.g., the image does not contain the point $(0,1,-1)$, which is in $C_2$).
Think of both $C_1$, $C_2$ as rotationally symmetric surfaces in $\mathbb R^3$, and write $r = \sqrt{x^2 + y^2}$. Then
$$\begin{split} C_1 &= \{ (r\cos\theta, r\sin \theta, z) : r =z \}, \\ C_2 &= \{(r\cos\theta , r\sin\theta, z) : r = 1\}. \end{split}$$ so if you ignore the $\theta$ coordinates for the moment, you need a diffeomorphism between the line $$ R_1 = \{ (r, r) : r >0 \}$$ and $$R_2 = \{ (1, s) :s \in \mathbb R\}.$$ Once such a diffeomorphism $ g : R_1 \to R_2$, $ (r, r)\mapsto g(r)$, is found, one can set $f : C_1 \to C_2$ as
$$ f(r\cos\theta, r\sin \theta, r) = (\cos\theta, \sin\theta , g(r)).$$