How many $f:\mathbb{R} \rightarrow \mathbb{R}$ are there satisfying $f(x)f(y)f(z)=12f(xyz)-16xyz$ for every real $x,y,z$?

$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 0 \qquad \textbf{(E)}\ \text{None}$
1. **Assume the form of the function:**
   Given the functional equation \( f(x)f(y)f(z) = 12f(xyz) - 16xyz \), we hypothesize that \( f(t) = kt \) for some constant \( k \). This is a reasonable assumption because the right-hand side involves the product \( xyz \).

2. **Substitute \( f(t) = kt \) into the equation:**
   \[
   f(x) = kx, \quad f(y) = ky, \quad f(z) = kz
   \]
   Substituting these into the given equation:
   \[
   (kx)(ky)(kz) = 12f(xyz) - 16xyz
   \]
   Simplifying the left-hand side:
   \[
   k^3xyz = 12f(xyz) - 16xyz
   \]

3. **Express \( f(xyz) \) in terms of \( k \):**
   Since \( f(xyz) = k(xyz) \), substitute this into the equation:
   \[
   k^3xyz = 12k(xyz) - 16xyz
   \]

4. **Factor out \( xyz \) from both sides:**
   \[
   k^3xyz = (12k - 16)xyz
   \]
   Since \( xyz \) is a common factor and can be non-zero, we can divide both sides by \( xyz \):
   \[
   k^3 = 12k - 16
   \]

5. **Solve the polynomial equation:**
   \[
   k^3 - 12k + 16 = 0
   \]
   To find the roots of this cubic equation, we can use the Rational Root Theorem, which suggests that possible rational roots are factors of the constant term (16) divided by factors of the leading coefficient (1). Thus, the possible rational roots are \( \pm 1, \pm 2, \pm 4, \pm 8, \pm 16 \).

6. **Test the possible rational roots:**
   - For \( k = 2 \):
     \[
     2^3 - 12 \cdot 2 + 16 = 8 - 24 + 16 = 0
     \]
     So, \( k = 2 \) is a root.
   - For \( k = -4 \):
     \[
     (-4)^3 - 12 \cdot (-4) + 16 = -64 + 48 + 16 = 0
     \]
     So, \( k = -4 \) is a root.

7. **Verify the roots:**
   The polynomial \( k^3 - 12k + 16 \) can be factored as:
   \[
   (k - 2)(k + 4)(k - 2) = 0
   \]
   This confirms that the roots are \( k = 2 \) and \( k = -4 \).

8. **Conclusion:**
   The functions \( f(t) = 2t \) and \( f(t) = -4t \) satisfy the given functional equation. Therefore, there are exactly two such functions.

The final answer is \(\boxed{2}\)