If $\mathrm{f}(1)=3, \mathrm{f}^{\prime}(1)=2$, then $\frac{\mathrm{d}}{\mathrm{d} x}\left\{\log \left[\mathrm{f}\left(\mathrm{e}^x+2 x\right)\right]\right\}$ at $x=0$ is

Given $y = \log\left[f\left(e^x + 2x\right)\right]$, the derivative $\frac{dy}{dx}$ is found using the chain rule. Let $u = f(e^x + 2x)$, so $y = \log u$ and $\frac{dy}{du} = \frac{1}{u}$.

Differentiating $u$ with respect to $v = e^x + 2x$ gives $\frac{du}{dv} = f'(v)$, while $\frac{dv}{dx} = e^x + 2$.

Combining these, $\frac{dy}{dx} = \frac{1}{f(e^x + 2x)} \cdot f'(e^x + 2x) \cdot (e^x + 2)$.

At $x = 0$, $e^0 + 2 \cdot 0 = 1$. Substituting $f(1) = 3$ and $f'(1) = 2$:

$\left.\frac{dy}{dx}\right|_{x=0} = \frac{1}{3} \cdot 2 \cdot (1 + 2) = 2$.

The derivative at $x = 0$ is $\boxed{2}$.