1. We need to evaluate

$$L=\lim_{x\to 0^+}\frac{\tan\left(5x^{1/3}\right)\,\ln(1+3x^2)}{\left(\tan^{-1}(3\sqrt{x})\right)^2\,\left(e^{5x^{4/3}}-1\right)}.$$

2. As $x\to 0^+$, all arguments go to $0$, so we use standard small-angle/small-value expansions:

• $\tan u \sim u$ as $u\to 0$
• $\ln(1+v) \sim v$ as $v\to 0$
• $\tan^{-1} w \sim w$ as $w\to 0$
• $e^z-1 \sim z$ as $z\to 0$

3. Apply these one by one:

• $\tan(5x^{1/3}) \sim 5x^{1/3}$
• $\ln(1+3x^2) \sim 3x^2$
• $\tan^{-1}(3\sqrt{x}) \sim 3\sqrt{x}$, so $\left(\tan^{-1}(3\sqrt{x})\right)^2 \sim (3\sqrt{x})^2=9x$
• $e^{5x^{4/3}}-1 \sim 5x^{4/3}$

4. Substitute into the limit:

$$L\sim \frac{(5x^{1/3})(3x^2)}{(9x)(5x^{4/3})}.$$

5. Simplify:

$$L\sim \frac{15x^{1/3+2}}{45x^{1+4/3}} =\frac{15x^{7/3}}{45x^{7/3}} =\frac{15}{45} =\frac{1}{3}.$$

6. Therefore,

$$\boxed{\frac{1}{3}}$$

which corresponds to Option C.

7. Comparison with stored answer:

• Stored correct answer: C
• Derived answer: C
• They match.