Finding Extrema of $f(x,y)=x^4+y^4-4xy$ Let $f(x,y)=x^4+y^4-4xy$ 

How do I find all the relative extrema and saddle points of $f$ which lie within the open square ${(x,y) | -2&lt;x&lt;2,-2&lt;y&lt;2}$. And also if $f$ was in the closed region ${(x,y) | -2\le x\le 2,-2\le y\le 2}$ what would be the absolute maxima and minima of $f$ on this closed region and where do these extreme values occur?

What I have Tried:

I first set $f_1=4(x^3-y)$ and $f_2=4(y^3-x)$ I don't know what to do next though.
Hint: Look at the corresponding closed region $R$, so that the extrema will either be critical points in int$R$ or they will lie on the boundary of $R$. For the saddle points, use the second derivative test.

Second hint: 

Step 1:

Solve the following equations and use the points in the second derivative test to decide if they are local extrema or saddle points. Suggestion: let $z=y/x$

$f_{x}=4x^{3}-4y=0\\f_{y}=4y^{3}-4x=0$

Step 2:

Now consider each piece of the boundary, making the appropriate substitution and extremize the function (of a single variable) you get in each case. For example, if we look at the top side of the square, where $y=2$ and $-2&lt;x&lt;2$, we get 

$f(x,2)=x^{4}-8x+16$, from which $f'(x,2)=4x^{3}-8=0$ gives $x=2^{1/3}$ as a critical point. Now determine what type of critical point it is, substitute into $f(x,y)$ and compare it with what you got in step 1. Then check $x=2, x=-2$ in $f(x,2)$ and write down the result. 

repeat for the other sides of the square.  

All told, you get a table of values from which you can read off the maxima and minima.
you can use Lagrangians ( Constrained Optimization). I give you an example here : 

http://mat.gsia.cmu.edu/classes/QUANT/NOTES/chap4.pdf