If f(x) is a differentiable function in 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 $f\left(\frac{3}{2}\right)$ is equal to : 

$$\begin{gathered} \underset{t\to x}{\lim } \frac{t^{2}f\left(x\right)-x^{2}f\left(t\right)}{t-x}=1 \\ \underset{t\to x}{\lim } \frac{2t f\left(x\right)-x^{2}f\text{'}\left(t\right)}{1}=1 \\ \Rightarrow 2xf\left(x\right)-x^2{f}^{\text{'}}\left(x\right)=1 \\ \Rightarrow \frac{dy}{dx}-\frac{2}{x}y=-\frac{1}{x^2} \\ IF=e^{\int -\frac{2}{x}dx}=\frac{1}{x^2} \end{gathered}$$
$$\begin{gathered} y IF=\int Q IF dx \\ \Rightarrow \frac{y}{x^2}=\int \frac{1}{x^4}dx \\ \Rightarrow \frac{y}{x^2}=-\frac{1}{3}\frac{1}{x^3}+c \\ \Rightarrow 1=-\frac{1}{3}+c\Rightarrow c=\frac{4}{3} \end{gathered}$$
$$\begin{gathered} y=-\frac{1}{3x}+\frac{4}{3}{x}^3 \\ \Rightarrow f\left(\frac{3}{2}\right)=-\frac{2}{9}+\frac{4}{3}\frac{27}{8} \\ =-\frac{2}{9}+\frac{9}{2}=\frac{77}{18} \end{gathered}$$