1. **State the problem:** Simplify the expression $\left(3x^2-\frac{1}{10}\right)\left(3x^2+\frac{1}{10}\right)$. 2. **Formula used:** This is a product of conjugates, which follows the difference of squares formula: $$ (a-b)(a+b) = a^2 - b^2 $$ where $a = 3x^2$ and $b = \frac{1}{10}$. 3. **Apply the formula:** $$ \left(3x^2-\frac{1}{10}\right)\left(3x^2+\frac{1}{10}\right) = (3x^2)^2 - \left(\frac{1}{10}\right)^2 $$ 4. **Calculate each square:** $$ (3x^2)^2 = 9x^4 $$ $$ \left(\frac{1}{10}\right)^2 = \frac{1}{100} $$ 5. **Substitute back:** $$ 9x^4 - \frac{1}{100} $$ 6. **Final answer:** $$ \boxed{9x^4 - \frac{1}{100}} $$