$\lim _{x \rightarrow 0} \frac{\mathrm{e}^{x^2}-\cos 3 x}{\sin x \log (1+2 x)}=$

The limit

$\lim_{x \to 0} \frac{e^{x^2} - \cos 3x}{\sin x \log (1 + 2x)}$

is of the indeterminate form $\frac{0}{0}$ as $x \to 0$.

Use standard approximations near zero: $e^u - 1 \approx u$, $1 - \cos u \approx \frac{u^2}{2}$, $\sin u \approx u$, and $\log(1 + u) \approx u$.

Rewrite the numerator as $(e^{x^2} - 1) - (\cos 3x - 1) = (e^{x^2} - 1) + (1 - \cos 3x) \approx x^2 + \frac{(3x)^2}{2} = x^2 + \frac{9x^2}{2} = \frac{11x^2}{2}$.
The denominator approximates to $(x)(2x) = 2x^2$.

Substitute into the limit:

$\lim_{x \to 0} \frac{\frac{11x^2}{2}}{2x^2} = \frac{11}{4}$

The limit evaluates to $\boxed{\frac{11}{4}}$.