Slope field of $y'=x^2 - y^2$ I don't know how I am supposed to go about creating a table with slope values for the graph so that I can sketch them. I knew how to do it when $y'$ equations had $y$ only or $x$ only, but not when both are in the equation.
At each point, make a tiny line segment with slope $x^2-y^2$.  For instance, at the point $(1,2)$, make a tiny line segment of slope $1^2-2^2=-3$.  A full graph is here:  http://www.wolframalpha.com/input/?i=slope+field+of+y%27%3Dx%5E2-y%5E2
You haven't provided much information but here are two possible ways of solving it.

Solution one:
This is assuming you know at least one point on the integrated graph.

$\frac{dy}{dx} = x^2 - y^2$

$y = \int x^2 - y^2 dx$

$ y = \frac{(x^3)}{3} + y^2x + c$

The following is where you should solve for c if you have your point, I am just going to carry it through

$xy^2 - y + (\frac{x^3}{3} + c) = 0$

You can either substitute into the quadratic formula here for y in terms of x.

$y = \frac{ 1 \pm\sqrt{1 - 4x(\frac{x^3}{3} + c})}{2x}$

Because your original equation ($ \frac{dy}{dy} = x^2 - y^2$) requires $y^2, $square both sides.

$y^2 = (\frac{ 1 \pm \sqrt{1-4x(\frac{x^3}{3} + c)}}{2x})^2$

Now you can substitute that in.

Solution two: This is assuming you know at least one point on the graph ($y'$).
I'm going to use $(a,b)$ as my point.

$b = a^2 + y^2$

$b - a^2 = y^2$

If you don't have either of the requirements for the solutions then I don't think you can graph that. (If I have made a mistake please point it out)