Let f(x) be continuously differentiable on the interval (0, $\infty$ )such that f(1) = 1, and $\underset{\mathrm{t}\to \mathrm{x}}{lim} \frac{{\mathrm{t}}^2\mathrm{f}\left(\mathrm{x}\right)-{\mathrm{x}}^2\mathrm{f}\left(\mathrm{t}\right)}{\mathrm{t}-\mathrm{x}}=1\text{ for each }\mathrm{x}>0\text{, }$ then f (x) is

$\underset{\mathrm{t}\to \mathrm{x}}{\mathrm{Lt}}\frac{{\mathrm{t}}^2\mathrm{f}\left(\mathrm{x}\right)-{\mathrm{x}}^2\mathrm{f}\left(\mathrm{t}\right)}{\mathrm{t}-\mathrm{x}}=1;\left(\frac{0}{0}\right)\text{ form }$;

$$\begin{gathered} \underset{\mathrm{t}\to \mathrm{x}}{\mathrm{Lt}}\frac{2\mathrm{tf}\left(\mathrm{x}\right)-{\mathrm{x}}^2{\mathrm{f}}^1\left(\mathrm{t}\right)}{1}=1; \\ {\mathrm{x}}^2{\mathrm{f}}^1\left(\mathrm{x}\right)-2\mathrm{xf}\left(\mathrm{x}\right)=-1 \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}-\frac{2}{\mathrm{x}}\mathrm{y}=-\frac{1}{{\mathrm{x}}^2}; \\ \text{ I.F= }{\mathrm{e}}^{\int -\frac{2}{\mathrm{x}}\mathrm{dx}}=1/{\mathrm{x}}^2\text{; } \end{gathered}$$

$\frac{\mathrm{y}}{{\mathrm{x}}^2}=\int \frac{-1}{{\mathrm{x}}^2}\cdot \frac{1}{{\mathrm{x}}^2}\mathrm{dx};\frac{\mathrm{y}}{{\mathrm{x}}^2}=\frac{1}{3{\mathrm{x}}^3}+\mathrm{c};\mathrm{f}\left(1\right)=1;$

$$\begin{gathered} 1=\frac{1}{3}+\mathrm{c}\Rightarrow \mathrm{c}=2/3; \\ \Rightarrow \mathrm{y}=\frac{1}{3\mathrm{x}}+\frac{2}{3}{\mathrm{x}}^2 \end{gathered}$$