I keep missing the factor of $t$. My working: $$(\vec x\cdot\nabla)\vec F=(x_i \vec e_i\cdot\vec e_j\frac{\partial}{\partial x_j})F_k\vec e_k=x_j\frac{\partial F_k}{\partial x_j}\vec e_k$$ and $$\frac{\partial\vec F}{\partial t}=\frac{\partial}{\partial t}\vec B(t\vec x)=(\vec x\cdot\nabla B_i(t\vec x))\vec e_i=(x_j\vec e_j\cdot\frac{\partial B_i(t\vec x)}{\partial x_k}\vec e_k)\vec e_i=x_k\frac{\partial B_i(t\vec x)}{\partial x_k}\vec e_i=x_k\frac{F_i}{x_k}\vec e_i$$ which would seem to suggest that the two lines are equal, when there should be a factor of $t$ between them.
I'm not sure I have differentiated correctly on the second line, but I can't see a better way.
What is the correct way to partial differentiate a time dependent vector field with respect to time?
I will work the problem where $F$ and $B$ are scalars and we have just the one position coordinate $x$. I leave it to you to generalize to the situation where $F$ and $B$ are time-dependent vector fields in three dimensions.
I think the best way to work such problems is the chain rule. Let $g(x,t) = tx$. Then $F = B \circ g$, so $$DF(x,t) = DB(g(x,t)) \cdot Dg(x,t) = DB(xt) \cdot Dg(x,t)$$ We have $$ Dg (x,t) = \begin{bmatrix} t & x \end{bmatrix} $$ Of course $$ DB(xt) = B'(xt) $$ So $$ DF(x,t) = \begin{bmatrix} tB'(xt) & xB'(xt)\end{bmatrix} $$ In particular, we see that $$ t \frac{\partial F}{\partial t}(x,t) = t \left(x B'(xt) \right) = tx B'(xt) $$ We also find $$ (x \cdot \nabla) F (x,t) = x \frac{\partial F}{\partial x} = x (t B'(xt)) = tx B'(xt) $$ Therefore, we conclude $$ (x \cdot \nabla) F = t \frac{\partial F}{\partial t} $$