$\begin{aligned} f(x) & =\frac{e^{2 x}-e^{-2 x}}{e^{3 x}+e^{-3 x}} \\ \Rightarrow f^{\prime}(x) & =\frac{\left(e^{3 x}+e^{-3 x}\right)\left(2 e^{2 x}+2 e^{-2 x}\right)}{\left(e^{2 x}-e^{-2 x}\right)\left(3 e^{3 x}-3 e^{-3 x}\right)} \\ \Rightarrow f^{\prime}(0) & =\frac{(1+1)(2 \cdot 1+2 \cdot 1)-(1-1)(3 \cdot 1-3 \cdot 1)}{(1+1)^2} \\ & =\frac{8}{4}=2 .\end{aligned}$