Let $f: \mathbb{R} \to \mathbb{R}$ a differentiable function.
If $f(0)>0$ and $f(x)(f(x)(1-x)+2xf'(x))=1, \forall x \in \mathbb{R}$ then
show that $f(0)=1$ and $f(x)=\sqrt{\frac{e^x-1}{x}}, x \neq 0$. Also show that the function $g=f^2$ is strictly increasing and convex in $\mathbb{R}$ and has the $y=0$-axis horizontal asymptotic in $-\infty$.
I have thought the following:
We have that $(x^2 f(x))'=2xf(x)+x^2 f'(x)$.
So $f(x)(1-x)+2xf'(x)=f(x)(1-x)-(x^2f(x))'-x^2 f'(x)$.
So we find other similar relations in order to get the desired result? Or do we proceed somehow else?
EDIT : We get the equation: $f^2(x)-xf^2(x)+2xf(x)f'(x)=1$.
By setting $g=f^2$ we get that $g'(x)+\frac{1-x}{x}g(x)=\frac{1}{x}$.
How do we solve this?