= Solution
In Cartesian coordinates on Euclidean space, define
$$
\boxed{\nabla_XY=\sum_{j=1}^n X(Y^j)\,\partial_j\qquad\text{when }Y=\sum_jY^j\partial_j.}
$$
This is an <affine connection> by the ordinary product rule. Its <connection coefficients> in this coordinate frame are all zero. The component formula for the <Lie bracket of vector fields> gives $\nabla_XY-\nabla_YX=[X,Y]$, so it is torsion-free. Also, since the Euclidean metric has constant coefficients,
$$
X\!\left(\sum_jY^jZ^j\right)=\sum_j X(Y^j)Z^j+\sum_jY^jX(Z^j).
$$
This is exactly metric compatibility. By uniqueness it is the Euclidean <Levi-Civita connection>. In particular constant vector fields are parallel, while the derivative of a general field is given by the boxed formula.
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