Let$f\left(x\right)$ be a differentiable on the interval

$(0,\mathrm{\infty })$such that $f\left(1\right)=1,$and $\underset{t\to x}{lim} \frac{t^{2}f\left(x\right)-x^{2}f\left(t\right)}{t-x}=1$for 

each x $> 0.$ Then $f\left(x\right)$is 

$1=\underset{t\to x}{lim} \frac{t^{2}f\left(x\right)-t^{2}f\left(t\right)+t^{2}f\left(t\right)-x^{2}f\left(t\right)}{t-x}$

$\begin{array}{l}=\underset{t\to x}{lim} \frac{t^2\left(f\right(x)-f(t\left)\right)}{t-x}+\underset{t\to x}{lim} (t+x)f\left(t\right)\\ =-x^2{f}^{\mathrm{\prime }}\left(x\right)+2xf\left(x\right)\end{array}$

Thus$f\left(x\right)$ is a solution of the differential equation

$x^2\frac{dy}{dx}-2xy=-1\Rightarrow \frac{dy}{dx}-\frac{2}{x}y=-\frac{1}{x^2}$     (1)

This is a linear equation with I.F. $=e^{-\int \frac{2}{x}dx}=\frac{1}{x^2}$

Multiplying (1) with $\left(1/x^2\right)$ we get 

$\frac{d}{dx}\left(\frac{y}{x^2}\right)=-\frac{1}{x^4}\Rightarrow \frac{y}{x^2}=\frac{1}{3x^3}+C$

Since $f\left(1\right)=1$so $C=\frac{2}{3}\therefore y=\frac{2}{3}{x}^2+\frac{1}{3x}$.