Maxima of bivariate function [1] Is there an easy way to formally prove that,
$$
2xy^{2} +2x^{2} y-2x^{2} y^{2} -4xy+x+y\ge -x^{4} -y^{4} +2x^{3} +2y^{3} -2x^{2} -2y^{2} +x+y$$
$${0<x,y<1}$$
without resorting to checking partial derivatives of the quotient formed by the two sides, and finding local maxima?
[2] Similarly, is there an easy way for finding $$\max_{0<x,y<1} [f(x,y)]$$
where,
$$f(x,y)=2x(1+x)+2y(1+y)-8xy-4(2xy^{2} +2x^{2} y-2x^{2} y^{2} -4xy+x+y)^{2}$$
 A: Certainly, there is no need for taking the quotient, since $a \ge b \Leftrightarrow \min \{a-b\} \ge 0$.
Here's a cool trick called the S.O.S. (sum of squares) method. The idea is to try and factor out $(x-y)^2$:
$$\begin{align}
LHS-RHS &=(x^4+y^4-2x^2y^2)-2(x^3+y^3-x^2y-xy^2)+2(x^2+y^2-2xy)\\
        &=(x^2-y^2)^2-2(x^2-y^2)(x-y)+2(x-y)^2\\
        &=(x-y)^2(x+y)^2-2(x-y)^2(x+y)+2(x-y)^2\\
        &=(x-y)^2((x+y)^2-2(x+y)+2)\\
        &=(x-y)^2((x+y-1)^2+1)\\
        &\ge 0
\end{align}$$
Note that this holds for all $x, y \in \mathbb R$.
A: Your first question:
With a little manipulation you get that it is equivalent to
$$x^2((1-x)^2+1)+y^2((1-y)^2+1) \ge 2xy[(1-x)(1-y)+1].$$
This can be obtained from addition of two inequalities
$$x^2(1-x)^2+y^2(1-y)^2 \ge 2xy(1-x)(1-y)$$
$$x^2+y^2\ge 2xy.$$
Both of them are special cases of $a^2+b^2\ge 2ab$, which follows from $(a-b)^2\ge 0$. (Or, if you prefer, you can consider it as a special case of AM-GM inequality.)
Note: To check the algebraic manipulations, you can simply compare the results for
2xy^2 +2x^2 y-2x^2 y^2 -4xy+x+y - ( -x^4 -y^4 +2x^3 +2y^3 -2x^2 -2y^2 +x+y)
expand x^2((1-x)^2+1)+y^2((1-y)^2+1) -2xy[(1-x)(1-y)+1]
Or simply subtract the two expressions:
2xy^2 +2x^2 y-2x^2 y^2 -4xy+x+y - ( -x^4 -y^4 +2x^3 +2y^3 -2x^2 -2y^2 +x+y) - [x^2((1-x)^2+1)+y^2((1-y)^2+1) -2xy[(1-x)(1-y)+1]]

I did not succeed in finding similar type of solution for your second problem.
