A nonzero polynomial $f(x)$ with real coefficients has the property that $f(x)=f^\prime(x)f^{\prime\prime}(x)$.  What is the leading coefficient of $f(x)$?
1. Given the polynomial \( f(x) \) with real coefficients satisfies the equation \( f(x) = f'(x) f''(x) \).
2. Let the degree of \( f(x) \) be \( n \). Then, the degree of \( f'(x) \) is \( n-1 \) and the degree of \( f''(x) \) is \( n-2 \).
3. The given equation implies:
   \[
   \text{deg } f = \text{deg } f' + \text{deg } f''
   \]
   Substituting the degrees, we get:
   \[
   n = (n-1) + (n-2)
   \]
4. Simplifying the equation:
   \[
   n = n - 1 + n - 2
   \]
   \[
   n = 2n - 3
   \]
   \[
   n - 2n = -3
   \]
   \[
   -n = -3
   \]
   \[
   n = 3
   \]
5. Therefore, the polynomial \( f(x) \) is of degree 3. Let the leading term of \( f(x) \) be \( ax^3 \).
6. The leading term of \( f'(x) \) is \( 3ax^2 \) and the leading term of \( f''(x) \) is \( 6ax \).
7. Substituting these into the given equation \( f(x) = f'(x) f''(x) \), we get:
   \[
   ax^3 = (3ax^2)(6ax)
   \]
8. Simplifying the right-hand side:
   \[
   ax^3 = 18a^2 x^3
   \]
9. Equating the coefficients of \( x^3 \) on both sides, we get:
   \[
   a = 18a^2
   \]
10. Solving for \( a \):
    \[
    a = 18a^2
    \]
    \[
    18a^2 - a = 0
    \]
    \[
    a(18a - 1) = 0
    \]
    Since \( a \neq 0 \), we have:
    \[
    18a - 1 = 0
    \]
    \[
    18a = 1
    \]
    \[
    a = \frac{1}{18}
    \]

The final answer is \(\boxed{\frac{1}{18}}\).