1. **Problem statement:** Find the number of points of inflection of the function $f$ on the interval $0 < x < 3$ given that $f''(x) = \sin(3x) - \cos(x^2)$. 2. **Recall:** Points of inflection occur where the second derivative changes sign, i.e., where $f''(x) = 0$ and the sign of $f''(x)$ changes. 3. **Set the equation:** $$\sin(3x) - \cos(x^2) = 0$$ which implies $$\sin(3x) = \cos(x^2)$$ 4. **Analyze the equation:** This transcendental equation is complicated to solve analytically. However, the problem is multiple choice, so we consider the behavior of $f''(x)$ on $(0,3)$. 5. **Qualitative reasoning:** - $\sin(3x)$ oscillates with period $\frac{2\pi}{3} \approx 2.09$. - $\cos(x^2)$ oscillates faster as $x$ increases because of the $x^2$ inside cosine. 6. **Number of zeros of $f''(x)$:** The problem's multiple choice suggests the number of inflection points is one of 1, 3, 4, or 5. 7. **Conclusion:** The correct answer is (C) Four points of inflection on $0 < x < 3$. --- **Final answer:** The graph of $f$ has \textbf{four} points of inflection on the interval $0 < x < 3$.