Find the critical numbers of the function. (Enter your answers as a comma-separated list.)
$$\begin{array}{l}f(x)=\frac{81x}{x^2+36}\\ x=◻\end{array}$$

Step 1 ：Given the function $f(x)=\frac{81x}{x^2+36}$, we are asked to find the critical numbers of the function. Critical numbers are where the derivative of the function is equal to zero or undefined.

Step 2 ：To find the derivative of the function, we need to use the quotient rule for differentiation. The quotient rule states that the derivative of a quotient u/v is $(v{u}^{\prime }-u{v}^{\prime })/v^2$, where u' and v' are the derivatives of u and v respectively.

Step 3 ：Applying the quotient rule, we find the derivative of the function to be $f^{\prime }(x)=-\frac{162x^2}{(x^2+36{)}^2}+\frac{81}{x^2+36}$.

Step 4 ：We then set the derivative equal to zero and solve for x to find the critical numbers. This gives us the critical numbers as -6 and 6.

Step 5 ：Final Answer: The critical numbers of the function are $\boxed{{\displaystyle -6,6}}$.