Compute the following limit. Use INF to denote $\infty$ and MINF to denote $-\infty$.
\[
\lim _{x \rightarrow 0} \frac{x}{\int_{x}^{x^{2}} \sqrt[3]{729-6 t^{3}} d t}=
\]

Step 1 ：First, we need to find the derivative of the numerator and the denominator.

Step 2 ：The derivative of the numerator, x, with respect to x is 1.

Step 3 ：The derivative of the denominator, which is an integral, is a bit more complex. By the Fundamental Theorem of Calculus, the derivative of an integral of a function from a to x, where a is a constant, is just the function evaluated at x. However, in this case, both the lower and upper limits of the integral are functions of x. Therefore, we need to use Leibniz's rule, which states that the derivative of an integral of a function from a(x) to b(x) is equal to the function evaluated at b(x) times the derivative of b(x) minus the function evaluated at a(x) times the derivative of a(x).

Step 4 ：So, the derivative of the integral from x to x^2 of \((729 - 6t^3)^(1/3)\) dt is: \(2x(729 - 6x^6)^(1/3) - (729 - 6x^3)^(1/3)\)

Step 5 ：Now, we can apply L'Hopital's rule: \[\lim _{x \rightarrow 0} \frac{x}{\int_{x}^{x^{2}} \sqrt[3]{729-6 t^{3}} d t} = \lim _{x \rightarrow 0} \frac{1}{2x(729 - 6x^6)^(1/3) - (729 - 6x^3)^(1/3)}\]

Step 6 ：As x approaches 0, the denominator becomes \(2*0*(729 - 0)^(1/3) - (729 - 0)^(1/3) = 0 - 729^(1/3) = -9\).

Step 7 ：Therefore, the limit is \(-1/9\).

Step 8 ：So, the final answer is \(\boxed{-1/9}\).