f(x,y)=1/3(x y) 1<=x,y<=2
f(x,y)=0 Elsewhere

and I would I like to know how you chose your limits (how you know what limits to use)

Do I understand the question correctly? Given the function f(x,y), you want to calculate the joint cumulative distribution function F(x,y)? I take that to mean the following:

F(x,y)=P(X \leq x,Y \leq y)

Therefore

F(x,y)=\int_{-\infty}^y\int_{-\infty}^x f(x',y')\;dx'\;dy'

The upper limits are x and y, since you're integrating (i.e. "adding up") all of the volume under the function for X and Y values less than x and y.

Given the function f(x,y) that you wrote, the function value is zero for all (x,y) < 1. Therefore you know that F(x,y) is zero for all (x<1) or (y<1). For the remaining cases (x,y \geq1), the lower limits of integration can be adjusted to 1.

Meanwhile, if x>2 or y>to the function value goes back to zero, so the upper limit of integration can be truncated at 2:

F(x,y)=0 for x<1 or y<1.

F(x,y)=\int_{1}^y\int_{1}^x \frac{1}{3}xy\;dx\;dy for 1<=x,y<=2.

F(x,y)=\int_{1}^y\int_{1}^2 \frac{1}{3}xy\;dx\;dy for x>2; 1<=y<=2.

F(x,y)=\int_{1}^2\int_{1}^x \frac{1}{3}xy\;dx\;dy for 1<=x<=2; y>2.

F(x,y)=\int_{1}^2\int_{1}^2 \frac{1}{3}xy\;dx\;dy for x,y>2.

Can you do the integrals?