Find, with explanation, the maximum value of $f(x)=x^3-3x$ on the set of all real numbers $x$ satisfying $x^4+36\leq 13x^2$.
1. **Identify the constraint and simplify it:**
   Given the constraint \( x^4 + 36 \leq 13x^2 \), we can rewrite it as:
   \[
   x^4 - 13x^2 + 36 \leq 0
   \]

2. **Solve the inequality:**
   To solve \( x^4 - 13x^2 + 36 \leq 0 \), we treat it as a quadratic in terms of \( y = x^2 \):
   \[
   y^2 - 13y + 36 \leq 0
   \]
   We solve the quadratic equation \( y^2 - 13y + 36 = 0 \) using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   y = \frac{13 \pm \sqrt{169 - 144}}{2} = \frac{13 \pm 5}{2}
   \]
   This gives us the roots:
   \[
   y = 9 \quad \text{and} \quad y = 4
   \]
   Therefore, the inequality \( y^2 - 13y + 36 \leq 0 \) can be factored as:
   \[
   (y - 9)(y - 4) \leq 0
   \]
   The solution to this inequality is:
   \[
   4 \leq y \leq 9
   \]
   Substituting back \( y = x^2 \), we get:
   \[
   4 \leq x^2 \leq 9
   \]
   Taking the square root of each part, we find:
   \[
   -3 \leq x \leq -2 \quad \text{or} \quad 2 \leq x \leq 3
   \]
   Thus, the interval for \( x \) is:
   \[
   x \in [-3, -2] \cup [2, 3]
   \]

3. **Find the critical points of \( f(x) = x^3 - 3x \):**
   To find the critical points, we take the derivative of \( f(x) \):
   \[
   f'(x) = 3x^2 - 3
   \]
   Setting the derivative equal to zero to find critical points:
   \[
   3x^2 - 3 = 0 \implies x^2 = 1 \implies x = \pm 1
   \]
   However, \( x = \pm 1 \) are not in the interval \( [-3, -2] \cup [2, 3] \).

4. **Evaluate \( f(x) \) at the endpoints of the interval:**
   We evaluate \( f(x) \) at \( x = -3, -2, 2, 3 \):
   \[
   f(-3) = (-3)^3 - 3(-3) = -27 + 9 = -18
   \]
   \[
   f(-2) = (-2)^3 - 3(-2) = -8 + 6 = -2
   \]
   \[
   f(2) = 2^3 - 3(2) = 8 - 6 = 2
   \]
   \[
   f(3) = 3^3 - 3(3) = 27 - 9 = 18
   \]

5. **Determine the maximum value:**
   Comparing the values:
   \[
   f(-3) = -18, \quad f(-2) = -2, \quad f(2) = 2, \quad f(3) = 18
   \]
   The maximum value is:
   \[
   \boxed{18}
   \]