Let $$f : R \to R$$ be defined as $$f(x) = \begin{cases} -\frac{4}{3}x^3 + 2x^2 + 3x, & x > 0 \\ 3xe^x, & x \le 0 \end{cases}$$
Then $$f$$ is increasing function in the interval

A function is increasing wherever its derivative is positive. We compute $$f'(x)$$ on each piece separately.

For $$x > 0$$: $$f(x) = -\frac{4}{3}x^3 + 2x^2 + 3x$$, so $$f'(x) = -4x^2 + 4x + 3$$. Setting $$f'(x) > 0$$ gives $$4x^2 - 4x - 3 < 0$$, which factors as $$(2x - 3)(2x + 1) < 0$$. This holds for $$-\frac{1}{2} < x < \frac{3}{2}$$. Since we are on the region $$x > 0$$, the derivative is positive for $$0 < x < \frac{3}{2}$$.

For $$x \le 0$$: $$f(x) = 3x e^x$$, so $$f'(x) = 3e^x + 3x e^x = 3e^x(1 + x)$$. Since $$e^x > 0$$ always, $$f'(x) > 0$$ when $$1 + x > 0$$, i.e., $$x > -1$$. So the derivative is positive for $$-1 < x \le 0$$.

We must also verify continuity at $$x = 0$$: from the left, $$f(0) = 0$$; from the right, $$\lim_{x \to 0^+} f(x) = 0$$. So $$f$$ is continuous. Combining the two intervals, $$f$$ is increasing on $$(-1, 0] \cup (0, \frac{3}{2}) = \left(-1, \frac{3}{2}\right)$$.

The answer is $$\left(-1, \frac{3}{2}\right)$$.