I have a hard time understanding how to integrate a double integral when I have a straight line and function like: $y=\frac{x}{3}$ and $x = y^2$ for the integral $f(x,y) = (x-y^2)$.
I evaluate it from there I have $y = \frac{x}{3}$ for $y$ upper limit and $y =\sqrt x$ for my lower limit and for $x = 9$ upper and $x = 0$ lower. And back on the question that thing I can't understand is on the first step $x-y^2$ I start to integrate with respect to $y$ now what the result should be after integration $xy -$ $y^3\over3$ is is this valid I dunno $x$ is constant then I multiply by $y$ and i get $xy$ and i integrate $y^3$ on the first step but I cannot get the right answer this way. I am interested only how the first step is done i know what to do after that. Than you for any help in advance.
Your boundaries are correct.
$$\int_0^9 dx \int_{\sqrt{x}}^{\frac{x}{3}} dy (x - y^2) = \int_0^9 dx \Big[xy - \frac{y^3}{3}\Big]_{\frac{x}{3}}^{\sqrt{x}} = \int_0^9 dx \Big( -\frac{x^2}{3} + \frac{x^3}{81} + \sqrt{x}x - \frac{\sqrt{x}^3}{81} \Big) $$
and the rest is straightforward calculation.
As you can you see, you first integrate to the variable whose borders are given in function of the other variable. By integrating, all instances of $y$ go away and get replaced by instances of $x$, leaving you with a one-dimensional integral.