Official solution
\[\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0\]
(Using L'Hospital's Rule)
\[ \Rightarrow \mathop {\lim }\limits_{t \to x} {{2t{f^2}(x) - 2{x^2}f(t).f'(t)} \over 1} = 0\]
\[ \Rightarrow \] 2xf2(x) - 2x2.f(x).f'(x) = 0
\[ \Rightarrow \] 2xf(x){f(x) - xf'(x)} = 0
[x \[ \ne \] 0, f(x) \[ \ne \] 0 as given
function \[f:\left( {0,\infty } \right) \to \left( {0,\infty } \right)\] only takes positive value as input and output]
\[ \Rightarrow f(x) = xf'(x) \]
\[\Rightarrow {{f'(x)} \over {f(x)}} = {1 \over x}\]
Integrating w.r.t x, we get
\[ \Rightarrow ln\,f(x) = ln\,x + ln\,C\]
\[ \Rightarrow f(x) = Cx\]
\[ \because \] f(1) = e
\[ \Rightarrow C = e;\,so\,f(x) = ex\]
When f(x) = 1 = ex
\[ \Rightarrow x = {1 \over e}\]