$f$ is an analytic function in the disk $D=\{z\in\mathbb{C}\,:\,|z|\leq 2\}$ such that $\iint_D=|f(z)|^2\,dx\,dy\leq 3\pi$. Maximize $|f''(0)|$ 
Determine the largest possible value of $|f''(0)|$ when $f$ is an analytic function in the disk $D=\{z\in\mathbb{C}\,:\,|z|<2\}$ with the property that $\iint_{D}|f(z)|^2\,dx\,dy\leq 3\pi$.

I don't really know what to do with the assumption that $\iint_D|f(z)|^2\,dx\,dy\leq 3\pi$. I believe you could use Stokes' theorem to rewrite this as a line integral on $\partial D$, but I'm really rusty with my usage, so I'm kind of stuck.
If I could get a bound on $\int_{\partial D}|f(z)|^2\,dz$, I could probably use harnack's inequality for subharmonic functions to get a bound on $|f|$ then Cauchy's inequality, however I'm not very sure about my usage of green's theorem (if that's even the right way to go) any help is greatly appreciated. Thanks

$$\langle f,f \rangle=\iint_Df(z)\times\overline{f(z)}dxdy=\iint_D |f(z)|^2dxdy \leqslant 3\pi$$
Power series of $f(z)$ about $z=0$:
$$\sum_{n=0}^{\infty}a_nz^n=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}z^n$$
then you can easily prove (using polar coordinates, edit: or more easily by defining a base using $z^n$ then using the generalized form of Parseval's Identity) that $$\langle f,f \rangle=\pi\sum_{n=0}^{\infty}\frac{|a_n|^2\times 2^{2n+2}}{n+1} \leqslant 3\pi$$
$$\pi \frac{|a_2|^2\times 2^{2*2+2}}{2+1} \leqslant \pi\sum_{n=0}^{\infty}\frac{|a_n|^2\times 2^{2n+2}}{n+1} \leqslant 3\pi$$
$$|f''(0)/2!|^2\times\frac{64}{3} \leqslant3$$
$$|f''(0)| \leqslant\frac{3}{4}$$
CAVEAT : there might be more than a few mistakes, it's been some time since I've last done something like this.
