I want to check my answer, pleas tell me if it's correct or not
first problem
$y=\left(\log _{\frac{1}{x}}\left(e\right)\right)$
my answer $y=\frac{lne}{ln_{\frac{1}{x}}}$
$y^|=lne\frac{d\left(\frac{1}{ln_{\frac{1}{x}}}\right)}{dx}=-\left(ln_{\frac{1}{x}}\right)^{-2}\frac{-x^{-2}}{\frac{1}{x}}=\left(lne\right)\left(\frac{-1}{\left(ln_{\frac{1}{x}}\right)^2}\right)\cdot \left(\frac{-1}{x}\right)$
$y^|=\frac{1}{x\left(ln_{\frac{1}{x}}\right)^2}$
second problem
$y=\log _{lnx}\left(e\right)$
my answer
$y=\frac{lne}{ln\left(lnx\right)}$ $y^|=y=lne\frac{d}{dx}\left(\frac{1}{ln\left(lnx\right)}\right)=\frac{\frac{-\frac{1}{xlnx}}{\left(ln\left(lnx\right)\right)^2}}{\left(ln\left(lnx\right)\right)^{-1}}=\frac{-\frac{1}{xlnx}}{ln\left(lnx\right)}$
$y^|==\frac{-1}{\left(xlnx\right)ln\left(lnx\right)}$
We have in general
$$y(x)=\log_{f(x)}e\implies y(x)\ln\left(f(x)\right)=1\implies y'(x)=-\frac{f'(x)}{f(x)\ln^2(f(x))} \tag 1$$
For $f(x)=1/x$, $(1)$ gives
$$y'(x)=\frac{1}{x\ln^2(x)}$$
while for $f(x)=\ln(x)$, $(1)$ gives
$$y'(x)=-\frac{1}{x\ln(x)\ln^2(\ln(x))}$$