1. Interpret the functions

The notation means $f(x)=2^{10}x+1, \qquad g(x)=3^{10}x-1.$

We are given $(f\circ g)(x)=x.$

That is, $f(g(x))=x.$

2. Compute $f(g(x))$

Since $g(x)=3^{10}x-1,$ we substitute into $f$: $f(g(x))=2^{10}(3^{10}x-1)+1.$

Now simplify: $f(g(x))=2^{10}3^{10}x-2^{10}+1.$

Since $2^{10}3^{10}=(2\cdot 3)^{10}=6^{10}$, we get $f(g(x))=6^{10}x-2^{10}+1.$

Given $f(g(x))=x$, so $6^{10}x-2^{10}+1=x.$

3. Solve for $x$

Bring $x$ terms together: $6^{10}x-x=2^{10}-1.$

Factor: $x(6^{10}-1)=2^{10}-1.$

Hence, $x=\frac{2^{10}-1}{6^{10}-1}.$

4. Rewrite to compare with options

Observe that $6^{10}=2^{10}3^{10}.$ So $x=\frac{2^{10}-1}{2^{10}3^{10}-1}.$

Now divide numerator and denominator by $3^{10}$: $x=\frac{\frac{2^{10}-1}{3^{10}}}{2^{10}-3^{-10}}.$

This is not directly one of the options. Instead, divide numerator and denominator by $2^{10}3^{10}$ or manipulate carefully:

Multiply option C denominator form: $\frac{1-3^{-10}}{2^{10}-3^{-10}}.$

Now note $1-3^{-10}=\frac{3^{10}-1}{3^{10}},$ so $\frac{1-3^{-10}}{2^{10}-3^{-10}}=\frac{\frac{3^{10}-1}{3^{10}}}{\frac{2^{10}3^{10}-1}{3^{10}}}=\frac{3^{10}-1}{2^{10}3^{10}-1}.$ This is not equal to our value.

Now check option B: $\frac{2^{10}-1}{2^{10}-3^{-10}}.$ Multiply numerator and denominator by $3^{10}$: $\frac{(2^{10}-1)3^{10}}{2^{10}3^{10}-1},$ again not equal.

Check option D: $\frac{1-2^{-10}}{3^{10}-2^{-10}}.$ Multiply numerator and denominator by $2^{10}$: $\frac{2^{10}-1}{2^{10}3^{10}-1}.$ This matches exactly.

Therefore, $x=\frac{1-2^{-10}}{3^{10}-2^{-10}}.$

5. Correct option

Thus the correct answer is Option D.