If $x+y=3e^2\text{then}\frac{d}{dx}\left(x^y\right)=0$ for x =

$y=3e^2-x\Rightarrow f\left(x\right)=x^y=x^{3e^2-x}$

$\Rightarrow \log f\left(x\right)=(3e^2-x)\log \left(x\right)\text{and diff w.r.t.}x$

$\frac{f\text{'}\left(x\right)}{f\left(x\right)}=\left(3e^2-x\right)\frac{1}{x}+\left(-1\right)\log x$

$f\text{'}\left(x\right)=f\left(x\right)\left(\left(3e^2-x\right)\frac{1}{x}+\left(-1\right)\log x\right)=0$

$\frac{3e^2}{x}-1-\log x=0\Leftarrow x\left(1+\log x\right)=3e^2$

$\therefore x=e^2$