This is how the problem goes: Let $T:\mathbb{R}^{n_1}\times...\times\mathbb{R}^{n_k}\rightarrow \mathbb{R}^m$ be multilinear. Prove $$DT(x_1,...,x_k)(y_1,...,y_k)=\sum_{i=1}^k T(x_1,...,y_i,...,x_k)$$
It can be easily proven that bilinear maps are differentiable and $DT(x_1,x_2)(y_1,y_2)=T(x_1,y_2)+T(y_1,x_2)$. However, I got stuck in trying to prove this inductively.
I tried constructing $T(x_1,...,x_k)=g(x_1,f(x_2,...,x_k))$ for some $g,f$ that are multilinear as well. So in the induction step, we can assume $$Df(x_2,...,x_k)(y_2,...,y_k)=\sum_{i=2}^k (x_2,...,y_i,...,x_k)$$ I also tried using chain rule, but that doesn't seem to go anywhere.
Can anyone help with where to go next?
Note that since $T$ is linear in each of the components, we have ${\partial T(x) \over \partial x_i}h_i = T(x_1,...,x_{i-1}, h_i, x_{i+1},...,x_n)$. Since $T$ is linear, the partial is continuous.
Hence $T$ is differentiable and
${\partial T(x) \over \partial x}h = \sum_i{\partial T(x) \over \partial x_i}h_i = \sum_i T(x_1,...,x_{i-1}, h_i, x_{i+1},...,x_n)$.
Alternative route:
If you want to go full monte, you could note that $\|T(x_1,...,x_k)\| \le L \|x_1\| \cdots \|x_k\|$ for some $L$. Then note that (skipping the inductive details) $T(x+h) = T(x) + \sum_i T(x_1,...,x_{i-1}, h_i, x_{i+1},...,x_n) + \sum_ {\text{terms with }\ge \ 2 h_i}T(\cdots)$. For $\|h\| \le 1$, we have $\|\sum_ {\text{terms with }\ge \ 2 h_i}T(\cdots) \| \le K \|h\|^2 $ for some $K$ that depends on $x$, $L$.
Hence $\|T(x+h)-T(x) - \sum_i T(x_1,...,x_{i-1}, h_i, x_{i+1},...,x_n)\| \le K \|h\|^2$, from which it follows that $T$ is Fréchet differentiable with derivative ${\partial T(x) \over \partial x}h = \sum_i T(x_1,...,x_{i-1}, h_i, x_{i+1},...,x_n)$.