Identifying a scalar function We know that a scalar is invariant under rotations. What about a scalar function? Should it also be invariant under rotations? Therefore, under rotation $\phi(x,y,z)$ must be equal to $\phi^\prime(x^\prime,y^\prime , z^\prime)$. Where $ (x^\prime,y^\prime , z^\prime)$ is the rotated coordinate system. Does it imply that $$\phi(x,y,z)=x^2+y^2+z^2$$ is the only possible scalar function in three dimension? Can $\phi(x,y,z)=x^2+yz$ be a scalar function?
 A: Good terminology question.
Let's work in some differentiable manifold $M$, our transformation is a smooth map $T: M \to N$. In the case of a rotation $M = N$.
Our $\phi$ is a smooth function $\phi: M \to \mathbb{R}$. 
In classical field theory the fact that $\phi$ maps to $\mathbb{R}$ is often expressed by the statement  "$\phi$ is a scalar field".
Now the additional demand that $\phi$ is in some sense invariant under $T$ (if $\phi$ is a field it is said that $T$ is a symmetry of the theory) is concretely the requirement:
$$T^*\phi |_{T(p)} = \phi |_p $$
for all $ p \in M$, where $*$ denotes the corresponding pullback on a smooth function. 
Physicist at times call this property "$\phi$ transforms like a scalar under $T$".
So to answer your last question:
The smooth map $\phi: (x, y, z) \mapsto x^2 + yz$:


*

*is a scalar valued function

*does not transform like a scalar under rotations.

A: As usual when dealing with transformations one has to be careful whether they are active or passive. If I understand your question you are implying a passive transformation, which is a mere change of coordinates. In this case all you are doing is changing the way you assign a scalar value to a "vector" of coordinates. Therefore $\phi(\mathbf x) = (\phi\circ\xi)(\mathbf x')$, where $\xi$ is the change of coordinates, i.e. $\mathbf x = \xi(\mathbf x')$. You can now set $\phi' = \phi\circ\xi$ to get the covariance for scalar fields
$$\phi(\mathbf x) = \phi'(\mathbf x').$$
To see how this relates to the more general case of tensor fields, where you have the Jacobian of the transformation changing components of tensors, consider that a pseudoscalar (like any tensor density of a certain weight) would transform with a certain power of the determinant of the Jacobian, 1 in this case, hence
$$\phi'(\mathbf x') = \det(D\xi)\phi(\mathbf x).$$
In the active picture you are actually taking the value of the scalar function from a point $p$ to a point $q$ of the manifold and requiring that $\phi(p) = \phi(q)$. This usually happens if $\phi$ has some sort of symmetry, like a spherical potential is invariant under rotation so that you can transport the scalar field along the flow of the transformation (this is well defined in the language of Lie derivatives along vector fields and indeed linked to the notion of symmetries).
