Does the cartesian unit vector x-hat depend on the spherical coordinate variables theta and phi?

No.

In order to understand why this seems like the wrong answer, let’s first start by defining our variables. Let’s use a Cartesian coordinate system where the x-axis is horizontal, the y-axis is vertical, and the z-axis is into/out of the page. Let’s further assume that the positive x-axis is to the right while the negative x-axis is to the left.

In this case, by definition, the unit vector \hat{x} always points to the right at any location in all of space. This is not normally the case for units vectors in other coordinate systems such as spherical (r,\theta,\phi). Let’s use the convention that \theta is the polar angle and \phi is the azimuthal angle. The direction that the unit vectors \hat{r}, \hat{\theta}, and \hat{\phi} points towards depends crucially on the specific location that is being considered. For example, \hat{r} always points radially away from the origin by definition. If you happen to be located on the x-axis at x = 1, then \hat{r} = +\hat{x}. On the other hand, if you were located on the y-axis at y = -2, then \hat{r} = -\hat{y}.

Now - here is where the confusion comes from.: the representation (not the definition) of \hat{x} in spherical coordinates is given by: \hat{x} = \sin \theta \cos \phi\ \hat{r} + \cos \theta \cos \phi \ \hat{\theta} - \sin \phi \ \hat{\phi}. This is an equation for \hat{x} which clearly shows that it depends on the variables \phi and \theta.

The tricky part here is that \hat{r}, \hat{\theta}, and \hat{\phi} all also secretly depend on \phi and \theta. Furthermore, \hat{r}, \hat{\theta}, and \hat{\phi} depend on \phi and \theta is just the right way to compensate or cancel out the dependence of \hat{x} on the sin and cos factors involving \phi and \theta.

In vector calculus, you often have to calculate integrals and derivatives involving unit vectors. If you forget that \hat{r}, \hat{\theta}, and \hat{\phi} have a \theta and a \phi dependence, you will get the wrong answer when integrating or taking the derivative with respect to the coordinate variables \theta and/or \phi. The Cartesian unit vectors \hat{x}, \hat{y}, and \hat{z} are truly constants and therefore independent of location. They can be safely ``moved outside’’ integrals and their derivatives are always zero.