Adding... it's easy, isn't it?

For $y$ constant, the sum of all $f(x,y)$ is $\sum _{ x=0 }^{ \infty }{ ( } \frac { 2x+3y }{ { 2 }^{ x+2y } } )=\quad 2\sum _{ x=0 }^{ \infty }{ ( } \frac { 2x+3y }{ { 2 }^{ x+2y } } )-\sum _{ x=0 }^{ \infty }{ ( } \frac { 2x+3y }{ { 2 }^{ x+2y } } )=\sum _{ x=0 }^{ \infty }{ ( } \frac { 2x+3y }{ { 2 }^{ x+2y-1 } } )-\sum _{ x=0 }^{ \infty }{ ( } \frac { 2x+3y }{ { 2 }^{ x+2y } } )=\frac { 3y }{ { 2 }^{ 2y-1 } } +\sum _{ x=1 }^{ \infty }{ ( } \frac { 2x+3y }{ { 2 }^{ x+2y-1 } } )-\sum _{ x=0 }^{ \infty }{ ( } \frac { 2x+3y }{ { 2 }^{ x+2y } } )=\frac { 3y }{ { 2 }^{ 2y-1 } } +\sum _{ x=0 }^{ \infty }{ \frac { 2 }{ { 2 }^{ x+2y } } } =\frac { 3y }{ { 2 }^{ 2y-1 } } +\sum _{ x=0 }^{ \infty }{ \frac { 1 }{ { 2 }^{ x+2y-1 } } } =\frac { 3y }{ { 2 }^{ 2y-1 } } +\sum _{ x=0 }^{ \infty }{ { 2 }^{ 1-2y-x } }$ .

Then, note that $\sum _{ n=k }^{ \infty }{ { 2 }^{ z-n } } ={ 2 }^{ z+1-k }$ .

In this case, $z=1-2y,\quad k=0$ so simplifying $\frac { 3y }{ { 2 }^{ 2y-1 } } +\sum _{ x=0 }^{ \infty }{ { 2 }^{ 1-2y-x } } =\frac { 3y }{ { 2 }^{ 2y-1 } } +{ 2 }^{ 2-2y }$ which is the sum of $f(x,y)$ for y constant and $x\ge 0$

Then the sum of all $f(x,y)$ for both $y,x\ge 0$ is $\sum _{ y=0 }^{ \infty }{ (\frac { 3y }{ { 2 }^{ 2y-1 } } +{ 2 }^{ 2-2y } } )$

Let ${ S }_{ 1 }=\sum _{ y=0 }^{ \infty }{ \frac { 3y }{ { 2 }^{ 2y-1 } } } =0+\frac { 3 }{ 2 } +\frac { 6 }{ 8 } +\frac { 9 }{ 32 } +\frac { 12 }{ 64 } +...$

${ S }_{ 2 }=\sum _{ y=0 }^{ \infty }{ { 2 }^{ 2-2y } } =4+1+\frac { 1 }{ 4 } +\frac { 1 }{ 16 } +\frac { 1 }{ 64 } +...$

Then,

$4{ S }_{ 1 }=0+6+\frac { 6 }{ 2 } +\frac { 9 }{ 8 } +\frac { 12 }{ 32 } +...\\ 4{ S }_{ 1 }-{ S }_{ 1 }=3{ S }_{ 1 }=6+\frac { 3 }{ 2 } +\frac { 3 }{ 8 } +\frac { 3 }{ 32 } +...\\ { S }_{ 1 }=2+\frac { 1 }{ 2 } +\frac { 1 }{ 8 } +\frac { 1 }{ 32 } +...\\ 4{ S }_{ 1 }=8+2+\frac { 1 }{ 8 } +...\\ 4{ S }_{ 1 }-{ S }_{ 1 }=3{ S }_{ 1 }=8\\ { S }_{ 1 }=\frac { 8 }{ 3 }$

Similarly,

$4{ S }_{ 2 }=16+4+1+\frac { 1 }{ 4 } +\frac { 1 }{ 16 } +...\\ 4{ S }_{ 2 }-{ S }_{ 2 }=3{ S }_{ 2 }=16\\ { S }_{ 2 }=\frac { 16 }{ 3 }$

Finally, ${ S }_{ 1 }+{ S }_{ 2 }=\frac { 24 }{ 3 } =\boxed{8}$