= Solution
The integrand and both limits depend on $x$, so use the <Leibniz integral rule>. The upper and lower endpoint contributions and the interior <derivative> give
$$
F'(x)=2x\sin(x^3)-3\sin(3x^2)+\int_{3x}^{x^2}t\cos(xt)\,dt.
$$
For $x\ne0$, an <antiderivative> in $t$ is $t\sin(xt)/x+\cos(xt)/x^2$. Evaluating it and collecting terms gives
$$
\boxed{F'(x)=3x\sin(x^3)-6\sin(3x^2)
+\frac{\cos(x^3)-\cos(3x^2)}{x^2}},\qquad x\ne0.
$$
At $x=0$ the original Leibniz expression has zero endpoint terms and an <integral> over a zero-length interval, giving $\boxed{F'(0)=0}$. The displayed quotient has a removable singularity and the same limiting value.
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