$\int \sec ^{\frac{2}{3}} x \cdot \operatorname{cosec}^{\frac{4}{3}} x \mathrm{~d} x=$

The integral $I = \int \sec^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x \mathrm{d}x$ can be rewritten as $I = \int \frac{\sec^2 x}{\tan^{\frac{4}{3}} x} \mathrm{d}x$ by expressing $\sec x = \frac{1}{\cos x}$ and $\operatorname{cosec} x = \frac{1}{\sin x}$, then multiplying numerator and denominator by $\sec^2 x$ to simplify.

Let $u = \tan x$, so $\mathrm{d}u = \sec^2 x \mathrm{d}x$. Substituting yields $I = \int u^{-\frac{4}{3}} \mathrm{d}u = -3 u^{-\frac{1}{3}} + c$.

Reversing the substitution gives $I = -3 \tan^{-\frac{1}{3}} x + c$. This matches option B.

Final Answer: $\boxed{\text{B}}$