Can anyone help me solve this differential equation? $y''+9y=0, y(\frac{\pi}{3})=3, y'(\frac{\pi}{3})=3$. I have got a general solution of $y(x)= C\cos(3x)+ D\sin(3x)$ but I am having trouble solving for the initial value portion
Your first initial condition gives you

$$y(\pi/3)=3=C \cos(3 \pi/3) + D \sin(3 \pi/3) = C \cos(\pi) + D \sin(\pi)=-C$$

so $C=-3$. Now differentiate and do the same thing for $y'(\pi/3)$.

In general you would need to solve a system of linear equations to deal with the initial conditions. Here we are fortunate that the equations nicely decouple, so we don't have to do that.
$y(\frac{\pi}{3}) = 3 \to -C = 0 \to C = 0, y'(\frac{\pi}{3}) = 3 \to -3D =3 \to D = -1$