= Solution
Assume $B$ is <continuously differentiable>. Apply the <chain rule> to each component with $z_j=tx_j$:
$$
\partial_{x_j}F_i=t\partial_{z_j}B_i(tx),\qquad
\partial_tF_i=x_j\partial_{z_j}B_i(tx).
$$
Contracting the first identity with $x_j$ proves
$$
\boxed{(x\cdot\nabla)F=t\,\partial_tF.}
$$
Here $\nabla$ differentiates with respect to $x$ at fixed $t$, while the <time derivative> holds $x$ fixed. The equality follows directly even at $t=0$; no division by $t$ is needed. It holds componentwise for the <vector field>, so it is not restricted to <scalar> functions.
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