To each pair of nonzero real numbers $a$ and $b$ a real number $a*b$ is assigned so that $a*(b*c) = (a*b)c$ and $a*a = 1$ for all $a,b,c$. Solve the equation $x*36 = 216$.
1. We start with the given properties of the operation $*$:
   \[
   a * (b * c) = (a * b) * c \quad \text{and} \quad a * a = 1 \quad \forall a, b, c \in \mathbb{R} \setminus \{0\}
   \]
   These properties suggest that $*$ is associative and that each element is its own inverse.

2. Let us denote the operation $*$ by a function $f(a, b)$ such that:
   \[
   f(a, f(b, c)) = f(f(a, b), c) \quad \text{and} \quad f(a, a) = 1
   \]

3. We need to solve the equation $x * 36 = 216$. This translates to finding $x$ such that:
   \[
   f(x, 36) = 216
   \]

4. To find the form of $f(a, b)$, we use the given properties. Consider the assertion $P(a, b, c)$:
   \[
   f(a, f(b, c)) = f(f(a, b), c)
   \]

5. Let's test $P(a, a, a)$:
   \[
   f(a, f(a, a)) = f(f(a, a), a)
   \]
   Since $f(a, a) = 1$, this simplifies to:
   \[
   f(a, 1) = f(1, a)
   \]
   This implies that $f(a, 1) = a$ because $f(1, a)$ should be $a$ for consistency.

6. Next, consider $P(a, b, b)$:
   \[
   f(a, f(b, b)) = f(f(a, b), b)
   \]
   Since $f(b, b) = 1$, this simplifies to:
   \[
   f(a, 1) = f(a, b) * b
   \]
   Given $f(a, 1) = a$, we have:
   \[
   a = f(a, b) * b
   \]
   Solving for $f(a, b)$, we get:
   \[
   f(a, b) = \frac{a}{b}
   \]

7. Now, we use this form of $f(a, b)$ to solve the original equation:
   \[
   f(x, 36) = 216 \implies \frac{x}{36} = 216
   \]
   Solving for $x$, we get:
   \[
   x = 216 \times 36 = 7776
   \]

The final answer is $\boxed{7776}$.