$f:(0,\infty )\to (0,\infty )\text{ be a differentiable function such that }f\left(1\right)=\mathrm{e}\text{ and }\underset{t\to x}{\lim }\frac{{\mathrm{t}}^2{f}^2\left(\mathrm{x}\right)-{\mathrm{x}}^2{f}^2\left(\mathrm{t}\right)}{\mathrm{t}-\mathrm{x}}=0$$\text{ If }f\left(\mathrm{x}\right)=1,\text{ then }\mathrm{x}\text{ is equal to: }$

$\begin{array}{l}\mathrm{f}:(0,\infty )\to (0,\infty ) \mathrm{f}\left(1\right)=\mathrm{e}\\ \underset{t\to x}{\lim }\frac{t^2{f}^2\left(x\right)-x^2{f}^2\left(t\right)}{t-x}\\ \text{ Using L' hospitals rule }\\ \underset{t\to x}{\lim }\frac{2{\mathrm{tf}}^2\left(\mathrm{x}\right)-{\mathrm{x}}^2·2\mathrm{f}\left(\mathrm{t}\right){\mathrm{f}}^{\text{'}}\left(\mathrm{t}\right)}{1}\\ =2{\mathrm{xf}}^2\left(\mathrm{x}\right)-{\mathrm{x}}^{2}2\mathrm{f}\left(\mathrm{x}\right){\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=0\\ \Rightarrow 2\mathrm{xf}\left(\mathrm{x}\right)=2{\mathrm{x}}^2{\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)\\ \frac{{\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}=\frac{1}{\mathrm{x}}\\ \text{ Integrate }\mathrm{lnf}\left(\mathrm{x}\right)=\mathrm{lnx}+\mathrm{C}\\ \mathrm{f}\left(1\right)=\mathrm{e}\Rightarrow \mathrm{C}=1\\ \mathrm{lnf}\left(\mathrm{x}\right)=\mathrm{lnx}+1\\ f\left(x\right)=ex\\ \text{ If }f\left(x\right)=1=ex\\ \Rightarrow x=\frac{1}{e}\end{array}$