Find the Global maximum and minimum values on the closed disk of this function just wondering if I am doing this correctly:
For $f(x,y) = x^3y + xy^3 - xy + 1 $ Find the Global maximum and minimum values on:
$ D = {(x,y) \in R  | x^2 + y^2 \le 4} $
I found that the critical points of the function are: $(0,0),(0,1),(0,-1),(\frac{1}{2},\frac{1}{2}),(\frac{1}{2},-\frac{1}{2}),(-\frac{1}{2},\frac{1}{2}),(-\frac{1}{2},-\frac{1}{2})$
Plugging each point into the function, I get:
$f(0,0) = 1, ), f(0,1) = 1, f(0,-1) = 1, f(\frac{1}{2},\frac{1}{2}) = \frac{7}{8}, f(\frac{1}{2},-\frac{1}{2}) = \frac{9}{8}, f(-\frac{1}{2},\frac{1}{2})  = \frac{9}{8}, f(-\frac{1}{2},-\frac{1}{2}) = \frac{7}{8}$
I then check the boundary points of D, by parametrising the variables, letting $x = 2cos\theta$ and $y = 2sin\theta$ for $ 0 \le \theta \le 2\pi$
Plugging these into the function and defining 
$g(\theta)= (2cos\theta)^3(2sin\theta)+ (2cos\theta)(2sin\theta)^3-(2cos\theta)(2sin\theta) + 1$
I take the derivative and get:
$g'(\theta) = 12 cos2\theta$, so $g$ has critical points at $\theta = \frac{\pi}{4} and \frac{5\pi}{4}$.
So the corresponding $x,y$ values are: $x=2cos(\frac{\pi}{4}) = \sqrt2, y=2sin(\frac{\pi}{4})=\sqrt2$ and $x=2cos(\frac{5\pi}{4}) = -\sqrt2, y=2sin(\frac{5\pi}{4})=-\sqrt2$
Finally, plugging these into $f$ I get:
$f(\sqrt2,\sqrt2) = (\sqrt2)^3(\sqrt2)+(\sqrt2)(\sqrt2)^3-(\sqrt2)(\sqrt2) +1 = 7$
$f(-\sqrt2,-\sqrt2) = (-\sqrt2)^3(-\sqrt2)+(-\sqrt2)(-\sqrt2)^3-(-\sqrt2)(-\sqrt2) +1 = 7$
So indeed I get max values at these points and min values for $x = \frac{1}{2}, -\frac{1}{2} $ and $y = \frac{1}{2}, -\frac{1}{2}$
Have I done this correctly? Many thanks for the feedback!

You may have worked too hard on the circle, and made a little error.
Our function is $xy(x^2+y^2)-xy+1$, which on the circle is $3xy+1$. This is $12\cos\theta\sin\theta+1$, which is $6\sin 2\theta+1$, maximum $7$, minimum $-5$. 
Note that parametrization (polar coordinates) could also be useful inside the circle, for our function is $\frac{1}{2}(r^4-r^2)\sin 2\theta +1$. The partial derivatives are easy to find. 
In finding critical points, rectangular coordinate version, there are some errors. For note the symmetry between $x$ and $y$. It follows that if for example $(0,1)$ is a critical point (which it is), then $(1,0)$ is also a critical point. 
But what's happening in the interior of the disk turns out to be irrelevant. The global maximum is $7$, and the global minimum is $-5$. Expressing the function as $xy(x^2+y^2-1)+1$ makes it clear that the global extrema must be on the boundary.
