= Solution
Put $\mathbf h=\mathbf y-\mathbf x$. For $0<t\le1$, <convexity> gives $f(\mathbf x+t\mathbf h)\le(1-t)f(\mathbf x)+tf(\mathbf y)$, hence
$$
f(\mathbf y)-f(\mathbf x)\ge\frac{f(\mathbf x+t\mathbf h)-f(\mathbf x)}{t}.
$$
By the first-order form of <Taylor's theorem>, the right side tends to $\mathbf h^{\mathsf T}\nabla f(\mathbf x)$ as $t\downarrow0$. Therefore
$$
\boxed{f(\mathbf y)\ge f(\mathbf x)+(\mathbf y-\mathbf x)^{\mathsf T}\nabla f(\mathbf x).}
$$
This is the <supporting inequality for a differentiable convex function>. The stated <partial derivatives> are enough: <coordinate derivatives give directional Taylor expansion for convex functions>, so no unmentioned continuity assumption on the second <partial derivatives> is needed. The result in fact needs less smoothness than the question assumes.
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