Reynolds transport theorem and eulerian material derivative

The problem as well as my work so far are in the image.

I begin with equation (2.23) and use the divergence theorem to express the second term as the surface integral part of the problem equation. From there I quickly end up with what appears to be an equation stating that the partial derivative of the scalar field at x is the same as the full derivative. This would imply that the integral of the convective part is zero. Since this is a general case I don't think I can make that statement. Either I've made a mistake somewhere or I am not seeing how to proceed from here.

Would appreciate help in figuring out which of those two it is and how I should proceed from here.

One thing that caught my eye repeatedly is the definition of the function for the problem. By definition and context I believe it is assumed that small x is always a function of time and initial position X, as in equation (2.20). Also, I do not fully understand why the second term of the final expression in (2.20) has to be evaluated at x.

Using 2.24, $$\begin{array}{l}\frac{d}{{dt}}\int_{CV}^{} {\phi d{\bf{x}} = } \int_{CV}^{} {\left( {\frac{{d\phi }}{{dt}} + \phi \frac{{\partial {v_i}}}{{\partial {x_i}}}} \right)d{\bf{x}}} \\ = \int_{CV}^{} {\frac{{d\phi }}{{dt}}d{\bf{x}}} + \int_{CV}^{} {\phi \frac{{\partial {v_i}}}{{\partial {x_i}}}d{\bf{x}}} \end{array}$$ From 2.20, $$= \int_{CV}^{} {\left( {\frac{{\partial \phi }}{{\partial {x_i}}}\frac{{\partial {x_i}}}{{\partial t}} + \frac{{\partial \phi }}{{\partial t}}} \right)d{\bf{x}}} + \int_{CV}^{} {\phi \frac{{\partial {v_i}}}{{\partial {x_i}}}d{\bf{x}}}$$ $$\begin{array}{l} = \int_{CV}^{} {\frac{{\partial \phi }}{{\partial {x_i}}}\frac{{\partial {x_i}}}{{\partial t}}d{\bf{x}}} + \int_{CV}^{} {\frac{{\partial \phi }}{{\partial t}}d{\bf{x}}} + \int_{CV}^{} {\phi \frac{{\partial {v_i}}}{{\partial {x_i}}}d{\bf{x}}} \\ = \int_{CV}^{} {\frac{{\partial \phi }}{{\partial {x_i}}}{v_i}d{\bf{x}}} + \int_{CV}^{} {\phi \frac{{\partial {v_i}}}{{\partial {x_i}}}d{\bf{x}}} + \int_{CV}^{} {\frac{{\partial \phi }}{{\partial t}}d{\bf{x}}} \\ = \int_{CV}^{} {\left( {\frac{{\partial \phi }}{{\partial {x_i}}}{v_i} + \phi \frac{{\partial {v_i}}}{{\partial {x_i}}}} \right)d{\bf{x}}} + \int_{CV}^{} {\frac{{\partial \phi }}{{\partial t}}d{\bf{x}}} \end{array}$$ From differentiation of multiplication, $$= \int_{CV}^{} {\frac{\partial }{{\partial {x_i}}}\left( {\phi {v_i}} \right)d{\bf{x}}} + \int_{CV}^{} {\frac{{\partial \phi }}{{\partial t}}d{\bf{x}}}$$ Finally, use the divergence theorem, $$= \int_{CS}^{} {\phi {v_i}d{\bf{s}}} + \int_{CV}^{} {\frac{{\partial \phi }}{{\partial t}}d{\bf{x}}}$$