Let $$f(x)=\sqrt{\lim _\limits{r \rightarrow x}\left\{\frac{2 r^2\left[(f(r))^2-f(x) f(r)\right]}{r^2-x^2}-r^3 e^{\frac{f(r)}{r}}\right\}}$$ be differentiable in $(-\infty, 0) \cup(0, \infty)$ and $f(1)=1$. Then the value of ea, such that $f(a)=0$, is equal to _________.

<p>$$\begin{aligned} & f(1)=1, f(a)=0 \\ & {f^2}(x) = \mathop {\lim }\limits_{r \to x} \left( {{{2{r^2}({f^2}(r) - f(x)f(r))} \over {{r^2} - {x^2}}} - {r^3}{e^{{{f(r)} \over r}}}} \right) \\ & = \mathop {\lim }\limits_{r \to x} \left( {{{2{r^2}f(r)} \over {r + x}}{{(f(r) - f(x))} \over {r - x}} - {r^3}{e^{{{f(r)} \over r}}}} \right) \\ & f^2(x)=\frac{2 x^2 f(x)}{2 x} f^{\prime}(x)-x^3 e^{\frac{f(x)}{x}} \\ & y^2=x y \frac{d y}{d x}-x^3 e^{\frac{y}{x}} \\ & \frac{y}{x}=\frac{d y}{d x}-\frac{x^2}{y} e^{\frac{y}{x}} \end{aligned}$$</p> <p>Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$</p> <p>$$\begin{aligned} & v=v+x \frac{d v}{d x}-\frac{x}{v} e^v \\ & \frac{d v}{d x}=\frac{e^v}{v} \Rightarrow e^{-v} v d v=d x \end{aligned}$$</p> <p>Integrating both side</p> <p>$$\begin{aligned} & \mathrm{e}^{\mathrm{v}}(\mathrm{x}+\mathrm{c})+1+\mathrm{v}=0 \\ & \mathrm{f}(1)=1 \Rightarrow \mathrm{x}=1, \mathrm{y}=1 \end{aligned}$$</p> <p>$$\begin{aligned} & \Rightarrow c=-1-\frac{2}{e} \\ & e^v\left(-1-\frac{2}{e}+x\right)+1+v=0 \\ & e^{\frac{y}{x}}\left(-1-\frac{2}{e}+x\right)+1+\frac{y}{x}=0 \\ & x=a, y=0 \Rightarrow a=\frac{2}{e} \\ & a e=2 \end{aligned}$$</p>