1. We need the term independent of $x$ in $\left(\frac{1}{60}-\frac{x^8}{81}\right)\left(2x^2-\frac{3}{x^2}\right)^6.$

2. First expand the general term of $\left(2x^2-\frac{3}{x^2}\right)^6.$

Using the binomial theorem, the general term is $T_{r+1}=\binom{6}{r}(2x^2)^{6-r}\left(-\frac{3}{x^2}\right)^r, \qquad r=0,1,2,\dots,6.$

Simplify: $T_{r+1}=\binom{6}{r}2^{6-r}(-3)^r x^{2(6-r)}x^{-2r}$ $=\binom{6}{r}2^{6-r}(-3)^r x^{12-4r}.$

So the power of $x$ in the general term is $12-4r.$

3. Now multiply by $\left(\frac{1}{60}-\frac{x^8}{81}\right).$ We want the total power of $x$ to be $0$.

There are two cases.

Case 1: From $\frac{1}{60}$

We need the term in the binomial expansion with power of $x$ equal to $0$: $12-4r=0 \implies r=3.$

For $r=3$, $T_4=\binom{6}{3}2^3(-3)^3x^0.$ Compute: $\binom{6}{3}=20,\quad 2^3=8,\quad (-3)^3=-27.$ So, $T_4=20\cdot 8\cdot (-27)=-4320.$

Multiplying by $\frac{1}{60}$ gives contribution $\frac{1}{60}(-4320)=-72.$

Case 2: From $-\frac{x^8}{81}$

Now we need the term in the binomial expansion with power of $x$ equal to $-8$ so that after multiplying by $x^8$, the total power becomes $0$: $12-4r=-8 \implies -4r=-20 \implies r=5.$

For $r=5$, $T_6=\binom{6}{5}2^1(-3)^5x^{-8}.$ Compute: $\binom{6}{5}=6,\quad 2^1=2,\quad (-3)^5=-243.$ Thus, $T_6=6\cdot 2\cdot (-243)x^{-8}=-2916x^{-8}.$

Multiplying by $-\frac{x^8}{81}$ gives contribution $-\frac{1}{81}\cdot (-2916)=36.$

4. Add both constant contributions: $-72+36=-36.$

5. Therefore, the term independent of $x$ is $\boxed{-36}.$

6. Comparing with the stored correct answer: stored answer is C, i.e. $-36$, which matches our result.