Is it possible that the axis of rotation of a planet also rotates? Assume that there is a planet that rotates around a axis. Is it possible that this axis also rotates around another axis?
For example, the planet is $\{(x, y, z)| x^2 + y^2 + z^2 \leq 1\}$ and this rotates around $\vec{OP} = (0, 0, 1)$ at first. Then the point $P$ rotates around $y$-axis, $P=(\sin{t}, 0, \cos{t})$.
 A: It sounds like you are referring to precession. The body rotates about a primary axis of rotation, but this axis also rotates around a second (precessionary) axis.
The Earth does exactly this. We rotate around our primary axis (which points roughly in the direction of the star Polaris) once per day. This axis itself sweeps round in the sky with a period of around 26,000 years.
A: A planet is a rigid body located in 3D space. 
Before addressing your question, its necessary to clarify some points about the relevant frame of reference. 
For a rigid body in 3D space, we require 6 coordinates to completely specify its position. Out of many choices, one convenient choice, is to consider 3 coordinates to specify a particular point on the body and the rest to determine the orientation of the body relative that point. The choice of a particular point on the body can be made by seeing if the rigid body is free from supports/hinges or not. If free, the center of mass of the body serves as a good choice. If not, then choose one of those hinge points as your reference point.
Let the reference point be O. You have two motions to consider: the motion of the center of mass and the rotational motion about the center of mass. Note that this is all there is to the description of an arbitrary motion.
You have two equations for the two motions:
$$ M \frac{d \mathbf{P}}{dt}=\mathbf{F}\tag{1}$$
$$\frac{d \mathbf{L}}{dt}=\mathbf{N}\tag{2}$$
where 
$$ \mathbf{P}=M \mathbf{V}\tag{3}$$
$$\mathbf{L}= \mathbf{I}.\mathbf{\omega}\tag{4}$$
(1) is the linear momentum theorem while (2) is the angular momentum theorem. $\mathbf{F}$ and $\mathbf{N}$ are the total force and total torque respectively while $\mathbf{P}$ and $\mathbf{L}$ are the linear momentum and angular momentum respectively. $\mathbf{I}$ and $\mathbf{\omega}$ are the inertia tensor and the angular velocity about the point O.
Now, notice that $\mathbf{P}$ is always parallel to $\mathbf{V}$ but $\mathbf{L}$ may not be parallel to $\mathbf{\omega}$. Also, $\mathbf{I}$ is not constant wrt axis fixed in space, but changes as the body rotates.
Now, coming back to your question, there can be two cases:
Case 1: 
For a planet under the influence of a single star, the $\mathbf{L}$ is conserved, which means:
$$\frac{d \mathbf{L}}{dt}=\mathbf{N}=0  \tag{5}$$
Now, let us choose a frame $\mathcal{F}$ attached to the planet. From (4) and (5), one can conclude that:
$$\mathbf{I}.\frac{d\mathbf{\omega}}{dt}+\mathbf{\omega}\times(\mathbf{I}.\mathbf{\omega})=0 \tag{6}$$ where all the quantities except the inertia tensor are referred to an inertial frame outside the planet(like the center of the star). The inertia tensor is wrt the axes in $\mathcal{F}$.
If you now choose the body axes as the principal axes(system of axes for the moment of inertia tensor is diagonal), (6) reduces to the following set of equations:
$$I_1\dot{\omega}_1+(I_3-I_2)\omega_3\omega_2=0\\
I_2\dot{\omega}_2+(I_1-I_3)\omega_1\omega_3=0\\
I_3\dot{\omega}_3+(I_2-I_1)\omega_2\omega_1=0$$
Since a sphere is symmetrical, $I_1=I_2=I_3$, which means 
$$\omega_i=\text{const} \; \forall \; i$$
So you have no precession as $\mathbf{\omega}$ is constant wrt say the center of the star.
Case 2:
If there are more astronomical bodies that graviationally interact with the planet, then $\mathbf{N}$ is not zero as the net force is no longer central($\mathbf{r}$ is not parallel to $\mathbf{F}$).
Then the relavant system of equations is:
$$I\dot{\omega}_1=N_1(t)\\
I\dot{\omega}_2=N_2(t)\\
I\dot{\omega}_3=N_3(t)$$
If furthermore some symmetry of the system allows $\mathbf{\omega}$ to be constant and have a constant angle wrt one of the principal axes of the body, then it will be identified as precessing about that body axis.
In any case, the change in the direction of $\mathbf{\omega}$ corresponds to the change in the direction of the rotation axis.
