Let $G$ be a Lie group and $X,Y\in T_eG$. Show that
$$[X,Y]=\left.\frac{\partial}{\partial s}\right\vert_{s=0}\left.\frac{\partial}{\partial t}\right\vert_{t=0}\exp(sX)\exp(tY)\exp(-sX)\exp(-tY).$$
No matter how I approach this I end up with the same wrong result.
By definition $$[X,Y]=\operatorname{ad}(X)Y=(T_e(\operatorname{Ad})X)Y$$
which in turn is
$$(T_e(\operatorname{Ad})X)Y=\left.\frac{\partial}{\partial s}\right\vert_{s=0}\operatorname{Ad}(\exp(sX))Y=\left.\frac{\partial}{\partial s}\right\vert_{s=0}\left.\frac{\partial}{\partial t}\right\vert_{t=0}\exp(\operatorname{Ad}(\exp(sX)tY))$$
and by a hint given on the sheet that's equal to $$\left.\frac{\partial}{\partial s}\right\vert_{s=0}\left.\frac{\partial}{\partial t}\right\vert_{t=0}\exp(sX)\exp(tY)\exp(-sX)$$
What's my mistake?
Edit: The exact text of the question is:
Let now $G$ be a Lie group. The commutator of two elements $x,y\in G$ is the element $c(x,y):=xyx^{-1}y^{-1}$. Show that for all $X,Y\in T_eG$ we have $$[X,Y]=\left.\frac\partial{\partial s}\right\vert_{s=0}\left.\frac\partial{\partial t}\right\vert_{t=0}c(\exp sX,\exp tY).$$ Hint: use relations like $x\exp Yx^{-1}=\exp(\operatorname{Ad}(x)Y)$.
(1) Here $\exp (X)$ is integral curve for $X$ at identity.
(2) If $F: M\times N\rightarrow L$ is smooth, then $$ \frac{d}{dt} F(c(t),d(t))= DF (c',d')=DF (c',0)+DF(0,d') $$ (cf. Lemma 5.3 in your lecture note or the proof of Lemma 5.2. That is $\frac{d}{dt}\bigg|_{t=0}\exp (tX)\exp (tY) =X+Y$
(3) $$\frac{d}{ds}\frac{d}{dt} \exp (sX)\exp(tY)\exp(-sX)\exp(-tY) = \frac{d}{ds}\frac{d}{dt} \exp \bigg(Ad_{\exp ( sX) } (tY) \bigg) \exp (-tY ) = \frac{d}{ds} [ Ad_{\exp (sX) }Y-Y] =[X,Y] $$