The equation of the tangent to the curve  $y=(2x-1)e^{2(1-x)}$ at the point of its maximum is 

$y^{\text{'}}\left(x\right)=-2(2x-1)e^{2(1-x)}+2e^{2(1-x)}=2e^{2(1-x)}$ $(-2 x+2)$

Thus $y^{\text{'}}\left(x\right)=0\Rightarrow x=1\text{. Since, }{y}^{\text{'}\text{'}}\left(1\right)<0\text{ so }(1,1)$ is

the point of maximum and an equation of tangent to the curve is $y-1=0(x-1),\text{ i.e. }y=1$