Let f: ( 0, ∞ ) → ( 0, ∞ ) be a differentiable function such that f...

$\lim_{t \to x}\dfrac{{t}^{2}{f}^{2} ( x ) - {x}^{2}{f}^{2} ( t )}{t - x} = 0$
Using L'Hospital

$\lim_{t \to x}\dfrac{2t{f}^{2} ( x ) - {x}^{2}2f ( t ) {f}^{'} ( t )}{1} = 0$
$⇒ 2x{f}^{2} ( x ) - {x}^{2}2f ( x ) {f}^{'} ( x ) = 0$
$⇒ 2xf ( x ) [ f ( x ) - x{f}^{'} ( x ) ] = 0$
But $f ( x ) \ne 0$

So, $x{f}^{'} ( x ) = f ( x )$
$⇒ x\dfrac{dy}{dx} = y$
$⇒ \dfrac{1}{y}dy = \dfrac{1}{x}dx$
Integration gives

$\ln y = \ln x + lnc$
$⇒ y = cx ⇒ f ( x ) = cx$
Now

$f ( 1 ) = c = e$ (given)

So, $f ( x ) = ex$
Now if $f ( x ) = 1$, then $ex = 1 ⇒ x = \dfrac{1}{e}$