$\underset{t\to x}{\lim }\frac{t^2{f}^2\left(x\right)-x^2{f}^2\left(t\right)}{t-x}=0$
Using L'Hospital

$\underset{t\to x}{\lim }\frac{2t{f}^2\left(x\right)-x^{2}2f\left(t\right)f^{\text{'}}\left(t\right)}{1}=0$
$\Rightarrow 2x{f}^2\left(x\right)-x^{2}2f\left(x\right)f^{\text{'}}\left(x\right)=0$
$\Rightarrow 2 x f\left(x\right)[ f(x)-x{f}^{\text{'}}(x\left)\right]=0$
But $f\left(x\right)\ne 0$

So, $x{f}^{\text{'}}\left(x\right)=f\left(x\right)$
$\Rightarrow x\frac{dy}{dx}=y$
$\Rightarrow \frac{1}{y}dy=\frac{1}{x}dx$
Integration gives

$\ln y=\ln x+\ln c$
$\Rightarrow y=cx\Rightarrow f\left(x\right)=cx$
Now

$f\left(1\right)=c=e$ (given)

So, $f\left(x\right)=e x$
Now if $f\left(x\right)=1$, then $ex=1\Rightarrow x=\frac{1}{e}$