I'm having a bit of trouble with the last bit of Problem 3.2 in Kirillov Jr.'s Introduction to Lie Groups and Lie Algebras.
(3.2) Let $f: \mathfrak{g} \rightarrow G$ be any smooth map such that $f(0)=1_G$ and $f_*(0) = \text{id}_{\mathfrak{g}}.$ Show that the group law in this coordinate system near the identity has the form $f(x) f(y) = f(x+y+B_f(x,y)+ \cdots )$ for some bilinear map $B: \mathfrak{g} \otimes \mathfrak{g} \rightarrow \mathfrak{g}$ and that $B_f(x,y)-B_f(y,x)= [x,y]$.
The inverse function theorem lets us write $f(x)\cdot f(y) = f(\mu_f(x,y))$ for some smooth map $\mu_f: \tilde{U} \rightarrow \mathfrak{g}$ defined on some open set $\tilde{U} \subset \mathfrak{g} \times \mathfrak{g}$. Restricting to $x=0$ then $y=0$ in the Taylor expansion of $\mu_f$ around $0 \times 0$ reveals that $\mu_f(x,y) = x+y+ B_f(x,y) + \cdots$ for some bilinear map $B_f : \mathfrak{g} \otimes \mathfrak{g} \rightarrow \mathfrak{g}$ which is defined everywhere even though $\mu_f$ might not be since we can rescale. We know that in the case where $f= \text{exp}: \mathfrak{g} \rightarrow G$ we have $B_{\text{exp}} (x,y)= \frac{1}{2} [x,y]$, which means it's enough to show that for any other $\tilde{f}$ satisfying the conditions of the problem statement we have $$B_f(x,y)-B_f(y,x)=B_{\tilde{f}}(x,y)-B_{\tilde{f}}(y,x)$$ for all $x,y \in \mathfrak{g}$. This is all I've got so far - I can't seem to find the right view to take in order to finish the problem.
Note that unlike the exponential map, in general we don't expect there to be a relationship between $f(x)$ and $f( \lambda x)$ for $\lambda \in \mathbb{R}$ since all we know is that $f$ is smooth.
Any help would be greatly appreciated!
Since $f(0)=0$ and $\frac{d}{dt}|_{t=0} f(tx)=f_*(x)=x$ for every $x\in\mathfrak{g}$, we have $$ \frac{d}{ds}{\large |}_{s=0}\frac{d}{dt}{\large |}_{t=0}\exp (sx)\exp (ty)\exp(-sx)= \frac{d}{ds}{\large |}_{s=0}\frac{d}{dt}{\large |}_{t=0}f(sx)f(ty)f(-sx)$$ for every $x,y\in\mathfrak{g}$. It follows from the group law that $$ f(sx)f(ty)f(-sx) = f(ty +ts (B_{f}(x,y) - B_{f}(y,x))-s^2 B_{f}(x,x)+\cdots),$$ so the right-hand side of the first equation is $B_f(x,y)-B_f(y,x)$. The same calculation for the left-hand side gives $[x,y]$ (indeed, this is the standard way to show that $\text{ad } x.y=[x,y]$, where $\text{ad}=\text{Ad}_*$ and $\text{Ad}$ is the adoint representation; see Lemma 3.15 in Kirllov's book).