Let $\underline{a}\in\mathbb R^n$ be a fixed vector. Find all functions $f:\mathbb R \to \mathbb R $ such that the divergence of the vector field $(f(\underline{a}\cdot \underline{x}))\underline{x}$, $x \in \mathbb R^n$ is everywhere equal to $1$.
Attempt:
I've begun by trying to work out what the vector field I've been given 'is'. So I wrote $$G=(f(\underline{a}\cdot \underline{x}))\underline{x}=(f(\underline{a}\cdot \underline{x}))(x_1,...,x_n)=(x_1f(\underline{a}\cdot\underline{x}),...,x_nf(\underline{a}\cdot\underline{x}))=(\alpha x_1,...,\alpha x_n)$$
and I know that given a vector field $\underline{v}(\underline{x})=(v_1(x_1,...,x_n),...,v_n(x_1,...,x_n))$ the divergence is $$\frac{\partial v_1}{\partial x_1}+...+\frac{\partial v_n}{\partial x_n}$$
The problem I have is that my vector field isn't of the correct form to work out the divergence so I guess this means I've done something wrong? Can anyone help?
There is nothing wrong. You must work a little bit more. In this case $$ v_i(x)=f(a\cdot x)\,x_i=f(a_1\,x_1+\dots+a_n\,x_n)\,x_i. $$ Now you have to calculate $\partial v_i/\partial x_i$, sum from $i=1$ to $i=n$ and equal the result to $1$.