I have this seemingly simple question that gives me hard time:
Subset $\Omega \subset \mathbb{R}^n$ is open and $f\in C^1(\Omega;\mathbb{R}^n)$. If $x,y\in\Omega$, where $x=(x_1,...,x_n)$, then we have following $$f(x)-f(y)=\int_0 ^1 \frac{\partial}{\partial t}f(y+t(x-y))dt= \int_0 ^1 \sum_{k=1}^{n}\frac{\partial}{\partial x_k}f(y+t(x-y))(x_k-y_k)dt$$ $$=\int_0 ^1 f'(y+t(x-y))(x-y)dt$$
My calculus is a bit rusty, so can you help explaining some of the previous steps?
Use the chain rule, let $z = y + t (x-y)$:
$$ \frac{\partial f(y + t (x-y))}{\partial t} = f'(z)\frac{\partial z}{\partial t} = f'(z)(x-y)$$
Since you are working in $\mathbb{R}^n$ then actually $f'(z) (x-y)$ means $(x-y)\cdot \nabla f(z)$