Big axis of an ellipse I drew a circle in this square and I transformed them (view in 3d).
How to find the angle between the big axis and $y$ or $x$ axis?
Blue plane rotated by 36° around  $y$ axis;
azimut is 30° and altitude is 20° in this example.
I look for a formula to find this angle.
(x'OX) is azimut $\theta$;
(ZOy') is elevation or altitude $\phi$;
(Zov) is $\alpha$
The observer is above the (OXY) plane and looks through the green plane.

 A: The main axis of the ellipse lies on the intersection of two planes: the blue plane that contains the circle and a plane perpendicular to the direction of view (for what you said, this is cylindrical projection) and passing through the center of the circle:


*

*the blue plane is perpendicular to the vector $\lt cos \alpha , 0, sin \alpha \gt $ where $\alpha $ is the angle rotated by the blue plane.

*the view plane is perpendicular to the projection direction - vector $\lt cos \theta cos \phi , sin \theta cos \phi, sin \phi \gt $ where $\theta $ is the azimut and $\phi $ is the elevation.
Knowing that both planes pass through the origin (recall that a plane ax+by+cz=0 is perpendicular to vector $\lt a,b,c \gt $ at the origin), you can compute the intersection line and hence the angles you want.
Note: I'm not sure of what is the axis convention you are using or the reference for the azimut, the drawing also is not clear about which is the point of view (from below? or the observer is behind?) You should take this into account to feed the right angles into the formulas.
EDIT: answering the comment below
for the blue plane, let's call it plane A:
$ x cos \alpha + z sin \alpha = 0 \rightarrow x = -z tan \alpha $ (1)
substituting in the eq. of view plane B:
$ x cos \theta cos \phi + y sin \theta cos \phi + z sin \phi = 0 $
you get
$ -z tan \alpha cos \theta cos \phi + y sin \theta cos \phi + z sin \phi = 0 $
dividing by  $ cos \phi$ and rearranging terms
$ \rightarrow z (tan \phi - tan \alpha cos \theta) + y sin \theta = 0 $ 
therefore $ y = z ( tan \alpha cos \theta -tan \phi )/sin \theta $ (2)
so your derivation is correct, this is the equation of a line contained in both planes A and B. In particular, any point that satisfies (1) is in the plane A. There is no error there, so maybe you are feeding the wrong $ \alpha $ angle. In all the previous framework it is assumed that:


*

*$ \alpha $ is the angle between the blue plane and the z axis (I assume here that y axis is in the plane)

*$ \theta $ is the angle between the x axis and the horizontal projection of the direction of view, or equivalently, the angle between the xz plane and the vertical plane containing the direction of view

*$ \phi $ is the angle between the direction of view and the horizontal xy plane.


As I said, I wasn't completely sure about where are your angles measured from. I took some assumptions based on your drawing and the values of the angles, but I could be perfectly wrong. That can explain the error, if we are using different conventions, but the algebra is good.
EDIT 2: what if we look from outside?

maybe I confused you because I didn't follow usual spherical coordinates (my bad).
NOTE: the dashed ellipse is the original circle BEFORE rotating it 36º (its position if $\alpha$ were 0 - assumed here it was originally in the yz plane). The plane B is actually not represented, but should be perpendicular to the red line. Note that the  actual position of this B plane, closer or further from the observer, is irrelevant in terms of projection, only matters its normal vector. But for simplicity is best placed at the origin.
