Let $$f : \mathbb{R} \to \mathbb{R}$$ be a thrice differentiable odd function satisfying $$f'(x) \geq 0$$, $$f''(x) = f(x)$$, $$f(0) = 0$$, $$f'(0) = 3$$. Then $$9f(\log_e 3)$$ is equal to _____.

We are given the differential equation $$f''(x)=f(x)$$ together with the facts that $$f$$ is thrice differentiable, odd, $$f'(x)\ge 0$$ for all $$x$$, $$f(0)=0$$ and $$f'(0)=3$$. Our goal is to find $$9f(\log_e 3)$$.

Step 1 - General solution of $$f''(x)=f(x)$$.
For the linear, constant-coefficient differential equation $$f''(x)=f(x)$$, the auxiliary equation is $$m^2-1=0$$ giving roots $$m=\pm1$$. Hence the general solution is
$$f(x)=C_1e^{x}+C_2e^{-x}$$ $$-(1)$$

Step 2 - Using the odd-function condition.
Because $$f$$ is odd, $$f(-x)=-f(x)$$ for every $$x$$. Substitute $$-x$$ in $$(1)$$:
$$f(-x)=C_1e^{-x}+C_2e^{x}$$.
Set this equal to $$-f(x)=-(C_1e^{x}+C_2e^{-x})$$:
$$C_1e^{-x}+C_2e^{x}= -C_1e^{x}-C_2e^{-x}$$.
Collecting like terms gives$$(C_1+C_2)(e^{x}+e^{-x})=0 \quad\text{for all }x.$$
Since $$e^{x}+e^{-x}\neq0$$, we must have $$C_1+C_2=0$$, so $$C_2=-C_1$$.

Step 3 - Rewrite $$f(x)$$ with a single constant.
Insert $$C_2=-C_1$$ in $$(1)$$:
$$f(x)=C_1(e^{x}-e^{-x})=2C_1\sinh x.$$
Let $$k=2C_1$$; then
$$f(x)=k\sinh x.$$ $$-(2)$$

Step 4 - Use the initial slope $$f'(0)=3$$.
Differentiate $$(2)$$:
$$f'(x)=k\cosh x.$$
At $$x=0$$, $$\cosh 0=1$$, so
$$f'(0)=k\cdot1=3\quad\Longrightarrow\quad k=3.$$

Step 5 - Explicit form of $$f(x)$$.
With $$k=3$$, equation $$(2)$$ becomes
$$f(x)=3\sinh x.$$
Notice $$f'(x)=3\cosh x\ge0$$ for all $$x$$, satisfying the given monotonicity condition.

Step 6 - Evaluate $$f(\log_e 3)$$.
Let $$u=\log_e 3$$, then $$e^{u}=3$$ and $$e^{-u}=1/3$$.
Recall $$\sinh u=\dfrac{e^{u}-e^{-u}}{2}$$, so
$$\sinh(\log_e 3)=\frac{3-\frac13}{2}=\frac{9-1}{6}=\frac83=\frac43.$$
Therefore
$$f(\log_e 3)=3\left(\frac43\right)=4.$$

Step 7 - Compute $$9f(\log_e 3)$$.
$$9f(\log_e 3)=9\times4=36.$$

Hence the required value is $$36$$.