Derivative of multivariable function $f: \mathbb{R}^n\setminus\{0\}\to \mathbb{R}^n, f(x)=g(\|x\|)x$
My attempt: I used the product and chain rule.
$$J_f(x)=x\cdot g'(\|x\|)\cdot \frac{1}{\|x\|}\cdot x+g(r)I_n$$
The solution says it's $J_f(x)=g'(r)\frac{1}{\|x\|}xx^T+g(r)I_n$
why do we have $xx^T$? Why the transpose?
You need to be careful when you write the dot where you really mean matrix multiplications. If $x$ is a column vector, $xx^T$ is an $n\times n$ matrix; but $x^Tx$ is not. In any case, $xx$ does not make sense and $x\cdot x$ would be a scalar if you do inner product.
You may work on the case when $n=2$ and write everything in terms of components to see what happens.