解：

已知\(\ln(1 + x) + \ln(1 - x)=x - \frac{1}{2}x^2 + o(x^2) - x - \frac{1}{2}x^2 + o(x^2)= - x^2 + o(x^2)\)

\(e^{2\sin x}=1 + 2\sin x + \frac{1}{2}(2\sin x)^2 + o(x^2)=1 + 2\sin x + 2\sin^2 x + o(x^2)\)

因此，\(-3 = \lim\limits_{x \to 0}\frac{xf(x) - e^{2\sin x} + 1}{-x^2}=\lim\limits_{x \to 0}\frac{xf(x) - (1 + 2\sin x + 2\sin^2 x + o(x^2)) + 1}{-x^2}\)

\(=\lim\limits_{x \to 0}\frac{xf(x) - 2\sin x}{-x^2} + \lim\limits_{x \to 0}\frac{-2\sin^2 x}{-x^2}\)

可以得出\(\lim\limits_{x \to 0}\frac{xf(x) - 2\sin x}{-x^2} = - 5\)，\(\lim\limits_{x \to 0}\frac{xf(x) - 2\left(x - \frac{1}{6}x^3 + o(x^3)\right)}{x^2} = 5\)，

进一步可以得出\(\lim\limits_{x \to 0}\frac{xf(x) - 2x}{x^2} = 5\)，以及\(\lim\limits_{x \to 0}\frac{f(x) - 2}{x} = 5\)，

\(\lim\limits_{x \to 0}[f(x) - 2] = 0\)，可得\(\lim\limits_{x \to 0}f(x) = 2 = f(0)\)，故\(f^\prime(0) = \lim\limits_{x \to 0}\frac{f(x) - 2}{x} = 5\) 。