It looks to me as if your text/prof whatever is being sloppy in describing surfaces. Surfaces in differential geometry are almost always parametric surfaces, i.e., we start with a function
$$
X : [a, b] \times [c, d] \to \Bbb R^3
$$
and look at the image $S = X([a, b] \times [c, d])$, which (if $X$ is nice enough), we call a "surface". Often $[a, b]$ and $[c, d]$ are both $[0, 1]$, but not always. To keep things neat, I'm going to use $U$ to denote $[a, b] \times [c, d]$, the domain of $X$>
For such a surface $S$, we say that $X$ is a "parameterization" of $S$. Associated to such a parameterization is a Gauss map, with the same domain:
$$
N : U \to \Bbb R^3 : (u, v) \mapsto \frac{X_u(u,v) \times X_v(u, v)}{||X_u(u, v) \times X_v(u, v)||}
$$
Some books choose instead to define $N$ as a function on $S$ itself, saying that if $p = X(u, v)$ for some particular $u,v$, then
$$N(p) = \frac{X_u(u,v) \times X_v(u, v)}{||X_u(u, v) \times X_v(u, v)||}
$$
In this form, we have
$$
N: S \to \Bbb S^2
$$
Notice the difference in the domains.
This isn't really quite as nice, because it's possible that $X$ is not one-to-one, but nonetheless this notation is pretty common.
In your case, to "compute the Gauss map", you'll want to do the second thing, and you can do it without even writing a parameterization of the cylinder: you want to associate to each point $(x, y, z)$ of the cylinder a unit normal vector. For instance at the point $(3, 0, 8)$, it's pretty easy to see that the normal points in the $x$ direction, so you can write down that
$$
N(3, 0, 8) = (1, 0, 0).
$$
You could also have written $(-1, 0, 0)$ -- without an explicit parameterization, it's impossible to decide which was meant! But you should at least make a choice that's continuous as you vary $x,y,z$.
The hypothesis says that $x(u,v)= f(u,v) N (u,v)$, where $f$ is a scalar, and $N$ the unit normal vector (if we fix the origin to be this point). We have to prove that $f$ is constant.
${\partial x\over \partial u }= {\partial f\over \partial u }N+ f {\partial N\over \partial u }$.
As $N$ is a unit vector, ${\partial N\over \partial u }$ is a tangential vector (derive $<N,N>=1$, as well as ${\partial x\over \partial u }$.
So computing the scalar product with $n$ we find ${\partial f\over \partial u }=0$, and similarly ${\partial f\over \partial v }=0$, whence the result.
Best Answer
Note that there is an ambiguity when talking about (normal) Gauss map, in particular when choosing the normal unit vector. Both your choice and the one given by the solution (which is just the negative of what you have) are both acceptable choices.
For example, if you use the chart $$ X(u, v) = (v, u, u^2+v^2),$$ you recover the second answer. If that answer uses $f(x, y,z) = z-x^2-y^2$ they recover what you have.