I use the 3 of Proposition 1.27 to compute equality above red line. $\nabla_X(dx^j\otimes\partial_i)= \nabla_X(dx^j)\otimes\partial_i+ dx^j\otimes\nabla_X(\partial_i)$.
Then, how to get the red line equality ?
And,How use $\nabla_X(tr(dx^j\otimes\partial_i))=0$ to get $\nabla_X(dx^j\otimes\partial_i)=0$ ?
It is not true that $\nabla_X({\rm tr}(dx^j\otimes\partial_i))=0$ implies $\nabla_X(dx^j\otimes\partial_i)=0$; indeed this latter equation is false for most coordinate systems. Remember that you don't need to show $$\nabla_X dx^j \otimes \partial_i = -dx^j \otimes \nabla_X \partial_i,$$ but only the contracted version
$$(\nabla_X dx^j) (\partial_i) = -dx^j (\nabla_X \partial_i).$$ (Remember that ${\rm tr} (\omega \otimes X) = \omega(X)$.)
Thus you can simply take the trace of your equation $$\nabla_X(dx^j\otimes\partial_i)= \nabla_X(dx^j)\otimes\partial_i+ dx^j\otimes\nabla_X(\partial_i)$$ and use property 4 to show the LHS is zero.