Theorem: Let $x_1, \dots, x_n$ be differentiable vector-valued functions. Then the multilinear function $M(x_1, ... ,x_n)$ is differentiable.
This is the proof: I know that $M$ is linear in $x_1, \dots, x_n$, so that if for all $x_i$ (for $i \neq j$) the function M is linear, then it is a linear function of $x_j$. But I can't understand the given proof.
By expanding along a basis, it is straightforward to show that $M$ is continuous.
If a function has continuous partials then it is differentiable. It is not hard to see that $M(x_1,...,x_k+h,...x_n) - M(x_1,...,x_k,...x_n) - M(x_1,...,h,...x_n) = 0$ so it follows that ${\partial M(x_1,...,x_n) \over \partial x_k}h = M(x_1,...,h,...x_n)$ and so is continuous. Hence $M$ is differentiable and
$DM((x_1,...,x_n))(h_1,...,h_n) = \sum_k {\partial M(x_1,...,x_n) \over \partial x_k}h_k$.