A sphere is given an initial push so that it starts rolling (without slipping) up an inclined plane. During its climb

Let the upward direction along the inclined plane be positive. The sphere rolls up the incline, so its velocity $v$ is positive and decreasing due to the component of gravity down the incline.

For rolling without slipping, the relation between linear velocity and angular velocity is $v = \omega R$. Since $v$ is decreasing, the angular velocity $\omega$ must also decrease.

To decrease the angular velocity, an angular deceleration $\alpha$ is required in the direction opposite to the rotation. The forces acting on the sphere are gravity, normal reaction, and friction.

Gravity and normal reaction pass through the center of mass and exert zero torque about it. The required torque can only be provided by the force of static friction at the point of contact.

To produce a torque that opposes the rotation, the static friction must act in the upward direction along the incline.

The work done by static friction during pure rolling is zero because the point of contact is instantaneously at rest.

The net force is directed down the incline since the sphere is decelerating, and the net torque is non-zero as there is an angular deceleration.

Answer: the direction of the force of friction on the sphere is up the incline.