If f(x) is a differentiable function in the interval (0, ∞) such...

Let $L = \lim_{t \to x} \dfrac{{t}^{2}f\left(x\right) - {x}^{2}f\left(t\right)}{t - x} = 1$

Applying L'Hospital Rule

$L = \lim_{t \to x} \dfrac{2tf\left(x\right) - {x}^{2}{f}^{′}\left(t\right)}{1} = 1$

Or $2xf\left(x\right) - {x}^{2}{f}^{′}\left(x\right) = 1 ⇒ \dfrac{dy}{dx} + \left(- \dfrac{2}{x}\right)y = - \dfrac{1}{{x}^{2}}$..........$( 1 )$

Solving the above linear differential equation we get

Integrating factor $= {e}^{∫ - \dfrac{2}{x}dx} = {e}^{- 2\ln x} = \dfrac{1}{{x}^{2}}$

After multiplying the equation \left(1\right)$by$\dfrac{1}{{x}^{2}}$, and simplifying the equation \left(1\right)$ becomes,

$\dfrac{d}{dx}\left(\dfrac{y}{{x}^{2}}\right) = - \dfrac{1}{{x}^{4}}$

$⇒ \dfrac{y}{{x}^{2}} = ∫ - \dfrac{1}{{x}^{4}}dx = \dfrac{1}{3{x}^{3}} + c$

$⇒ y = \dfrac{1}{3x} + c{x}^{2}$

Putting $x = 1, y = 1$, we get

$c = \dfrac{2}{3}$.

$∴ f\left(x\right) = \dfrac{2}{3}{x}^{2} + \dfrac{1}{3x}$

Put $x = \dfrac{3}{2}$

$f\left(\dfrac{3}{2}\right) = \dfrac{2}{3} \times {\left(\dfrac{3}{2}\right)}^{2} + \dfrac{2}{3 \times 3}$

$= \dfrac{3}{2} + \dfrac{2}{9} = \dfrac{27 + 4}{18} = \dfrac{31}{18}$.