As always, to find the maximum and minimum of a function, you have to calculate first the critical points, that is
$$\frac{\partial G(x,y)}{\partial x}=0$$
$$\frac{\partial G(x,y)}{\partial y}=0$$
In this case, the respective equations are
$$\frac{\partial G(x,y)}{\partial x}=-\frac{1}{x^{2}}+\frac{9}{(4-x-y)^{2}}$$
$$\frac{\partial G(x,y)}{\partial y}=-\frac{4}{y^{2}}+\frac{9}{(4-x-y)^{2}}$$
Simplifying you should obtain $x=1-\displaystyle \frac{y}{4}$ and $\displaystyle \frac{5y}{2}=4-x$ which lead to the solution $(x,y)=\bigg\{\displaystyle \frac{2}{3},\displaystyle \frac{4}{3}\bigg\}$
Then you substitute those values in your original function $G(\frac{2}{3},\frac{4}{3})=9$.
As I said, such a point is just a critical point (which means, it could be a local minimum, a local maximum or a saddle point). In order to figure out, we have to use the Second partial derivative test.
You can see the details in the Wikipedia article. As you already know what you want to prove (that $(\frac{2}{3},\frac{4}{3})$ is a local minimum), we expect that $M(\frac{2}{3},\frac{4}{3})>0$ and $f_{xx}(\frac{2}{3},\frac{4}{3})>0$.
I think I'll leave it there. You can do the rest of the calculations.
By the way, the aforementioned procedure works to obtain only a local minimum, maximum or saddle point, not global extrema.