= Solution
The <total derivative> of $f$ at $x$ is a <linear map> $Df(x):\mathbb R^n\to\mathbb R$ satisfying
$$
\boxed{\lim_{h\to0}\frac{|f(x+h)-f(x)-Df(x)h|}{\|h\|}=0.}
$$
The <gradient> is the <vector> of <partial derivatives>, $\nabla f(x)=(\partial_1f(x),\ldots,\partial_nf(x))$. When $f$ is totally differentiable, these <partial derivatives> exist and
$$
\boxed{Df(x)h=\nabla f(x)\cdot h=\sum_{j=1}^n\partial_jf(x)h_j.}
$$
Thus the derivative is a linear functional represented by the <gradient> under the <Euclidean inner product>. Existence of <partial derivatives> alone does not imply total differentiability.
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