Let's solve the problem.

We are given: $(f \circ g)(x) = 9x^2 - 6x - 3$ and $g(x) = 3x - 1.$ We need to find the value of $f^{-1}(12)$.

Step 1: Determine $f(x)$

Since $(f \circ g)(x) = f(g(x))$, we substitute $g(x) = 3x - 1$ into $f(g(x)) = 9x^2 - 6x - 3$ to find $f(x)$.

Let $u = g(x) = 3x - 1$. Then, $f(u) = 9x^2 - 6x - 3.$

To express $f$ in terms of $u$, solve $x$ in terms of $u$ from $u = 3x - 1$: $x = \frac{u + 1}{3}.$ Substitute this expression for $x$ back into $f(g(x))$: $f(u) = 9\left(\frac{u + 1}{3}\right)^2 - 6\left(\frac{u + 1}{3}\right) - 3.$

Expanding and simplifying gives $f(u)$.

Step 2: Find $f^{-1}(12)$

Once