[tex]y=\dfrac{e^{2x}}{4+3e^x}[/tex]
[tex]\implies y'=\dfrac{2e^{2x}(4+3e^x)-3e^xe^{2x}}{(4+3e^x)^2}[/tex]
[tex]\implies y'=\dfrac{8e^{2x}+3e^{3x}}{(4+3e^x)^2}[/tex]
Stationary points occur where the derivative is zero. The denominator is positive for all [tex]x[/tex], so we only need to worry about the numerator.
[tex]8e^{2x}+3e^{3x}=e^{2x}(8+3e^x)=0[/tex]
[tex]e^{2x}>0[/tex] for all [tex]x[/tex], so we can divide through:
[tex]8+3e^x=0\implies e^x=-\dfrac83[/tex]
But [tex]e^x>0[/tex] for all [tex]x\in\mathbb R[/tex], so this function has no stationary points...
I suspect there may be a typo in the question.
