Let f(x) = _ r → x \ 2 r ^ 2 [( f ( r ) ) ^ 2 - f ( x ) f ( r )] r...

Given,

$f ( 1 ) = 1, f ( a ) = 0$

And $f\left(x\right) = \sqrt{\lim_{r \to x}\left\{\dfrac{2{r}^{2}\left[( f ( r ) {)}^{2} - f ( x ) f ( r )\right]}{{r}^{2} - {x}^{2}} - {r}^{3}e\dfrac{f ( r )}{r}\right\}}$

$⇒ {f}^{2} ( x ) = \lim_{r \to x}\left(\dfrac{2{r}^{2}\left({f}^{2} ( r ) - f ( x ) f ( r )\right)}{{r}^{2} - {x}^{2}} - {r}^{3}e\dfrac{f ( r )}{r}\right)$

$⇒ {f}^{2} ( x ) = \lim_{r \to x}\left(\dfrac{2{r}^{2}f ( r )}{r + x}\dfrac{( f ( r ) - f ( x ) )}{r - x} - {r}^{3}e\dfrac{f ( r )}{r}\right)$

$⇒ {f}^{2} ( x ) = \dfrac{2{x}^{2}f ( x )}{x + x}\left(\lim_{r \to x}\dfrac{( f ( r ) - f ( x ) )}{r - x}\right) - {x}^{3}e\dfrac{f ( x )}{x}$

$⇒ {f}^{2} ( x ) = \dfrac{2{x}^{2}f ( x )}{2x}{f}^{'} ( x ) - {x}^{3}e\dfrac{f ( x )}{x}$

Now, taking $f\left(x\right) = y$ we get,

$⇒ {y}^{2} = xy\dfrac{dy}{dx} - {x}^{3}e\dfrac{y}{x}$

$⇒ \dfrac{y}{x} = \dfrac{dy}{dx} - \dfrac{{x}^{2}}{y}e\dfrac{y}{x}$

Now, let $y = vx ⇒ \dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$

$⇒ v = v + x\dfrac{dv}{dx} - \dfrac{x}{v}{e}^{v}$

$⇒ \dfrac{dv}{dx} = \dfrac{{e}^{v}}{v}$

$⇒ {e}^{- v}vdv = dx$

Integrating both side

${e}^{v} ( x + c ) + 1 + v = 0$

Now using $f ( 1 ) = 1 ⇒ x = 1, y = 1$ we get,

$⇒ c = - 1 - \dfrac{2}{e}$

Hence, ${e}^{v}\left(- 1 - \dfrac{2}{e} + x\right) + 1 + v = 0$

$⇒ e\dfrac{y}{x}\left(- 1 - \dfrac{2}{e} + x\right) + 1 + \dfrac{y}{x} = 0$

Now taking, $x = a \text{ and } y = 0$ we get,

${e}^{0}\left(- 1 - \dfrac{2}{e} + a\right) + 1 + 0 = 0$

$⇒ a = \dfrac{2}{e}$

$⇒ ae = 2$