A cylinder like the one you have given is basically obtained by revolving a straight line at a fixed distance from a coordinate axis around that axis itself(z-axis in this case). So at each point of the circle of revolution, you have a sraight line passing through. Just take a ruled piece of rectangular paper and roll it up so that the ruled lines will give you the rulings which are straight lines.
The cross sectional curves are circles. So the cylinder has a corresponding tangent plane at each point of intersection of the straight line rulings and the cross sectional circles.It is spanned by two tangent vectors. One tangent $(e_1)$ would be along the straight line itself, other would be $(e_2)$ tangential to the circle.
Now how would you look at the normal to a circle??That would be how the normal to a cylinder would look like too, pointing away from the circle's centre. This normal would be perpendicular to the plane that is spanned by the two tangent vectors $e_1,e_2$.
Explicitly just look at the gradient to the function giving the cylinder equation and compute it at each point. That will give you the normal vector. Of course you need to plug in the constraints.
P.S.: The outward direction is because that is the direction in which the value of the function increases.