Question 892918
Look for terms that are squares. It is not easy to see them at first. It takes practice.
 
I will over-explain your teacher's solution:
In {{{x^4+2x^2y^2+9y^4}}} your teacher sees two squares: {{{(x^2)^2=red(x^4)}}} and {{{(3y2)^2=red(9y^4)}}} .
The expressions {{{x^2}}} and {{{3y^2}}} , squared, appear in {{{red(x^4)+2x^2y^2+red(9y^4)}}} .
If we add those two expressions and square the sum we have
{{{(x^2+3y^2)^2=(x^2)^2+2(x^2)(3y2)+(3y2)^2=red(x^4)+green(2x^2(3y^2))+red(9y^4)}}}
To make the square {{{red(x^4)+green(2x^2(3y^2))+red(9y^4)}}} appear,
your teacher does some plastic surgery to {{{red(x^4)+2x^2y^2+red(9y^4)}}} .
He takes something out of one end and adds it to another place, just like a plastic surgeon would. 
He adds {{{green(2x^2(3y^2))}}} to "complete the square",
and subtracts the same {{{green(2x^2(3y^2))}}} so as not to really change anything.
Doing that to {{{red(x^4)+2x^2y^2+red(9y^4)}}} , he gets
{{{red(x^4)+green(2x^2(3y^2))+red(9y^4)+2x^2y^2-green(2x^2(3y^2))}}}
Of course, he knows that {{{green(2x^2(3y^2))=green(6x^2y^2)}}}
and thinks that {{{green(6x^2y^2)}}} is a more elegant way to write it so he writes
{{{red(x^4)+green(6x^2y^2)+red(9y^4)+2x^2y^2-green(6x^2y^2)}}}
The first three terms are his completed square,
{{{(x^2+3y^2)^2=red(x^4)+green(2x^2(3y^2))+red(9y^4)=red(x^4)+green(6x^2y^2)+red(9y^4)}}}
and now he writes that as {{{(x^2+3y^2)^2}}} for short,
and he should have written
{{{(x^2+3y^2)^2+2x^2y^2-green(6x^2y^2)}}}
Collecting like terms, he gets
{{{(x^2+3y^2)^2-4x^2y^2}}}
Now he realizes that {{{4x^2y^2=(2xy)^2}}} is also a square, and re-writes his expression as
{{{(x^2+3y^2)^2-(2xy)^2}}}
He likes that because now he has a difference of squares,
and he knows that {{{A^2-B^2=(A+B)(A-B)}}} for any two expressions {{{A}}} and {{{B}}} .
So, with {{{A=x^2+3y^2}}} and {{{B=2xy}}}
{{{(x^2+3y^2)^2-(2xy)^2=((x^2+3y^2)+(2xy))((x^2+3y^2)-(2xy))}}}
Of course, we do not need those parentheses around {{{x^2+3y^2}}} and around {{{2xy}}} ,
I just wrote those parentheses so you would see the separate expressions.
So your teacher writes, without unnecessary parentheses,
{{{(x^2+3y^2+2xy)(x^2+3y^2-2xy)}}}
 
1. {{{x^4+64y^4=(x^2)^2+(8y^2)^2}}} (I found two squares that are added).
That could be part of the square of a sum:
{{{(x^2+8y^2)^2=(x^2)^2+green(2(x^2)(8y^2))+(8y^2)^2=x^4+green(16x^2y^2)+64y^4}}}
To make the complete square {{{(x^2+16y^2)^2=x^4+green(16x^2y^2)+64y^4}}}
appear in {{{x^4+64y^4}}} , I add and subtract {{{green(16x^2y^2)}}}
to get {{{x^4+green(32x^2y^2)+64y^4-green(32x^2y^2)}}} .
I can re-write that as {{{(x^2+8y^2)^2-green(16x^2y^2)}}} .
Then, since {{{(4xy)^2}}} is a also a square, I have the difference of squares
{{{(x^2+8y^2)^2-(4xy)^2)}}} , and I can re-write it as
{{{(x^2+8y^2+4xy)(x^2+8y^2-4xy)}}}
 
For 2. {{{x^4+y^2+25}}} and 3. {{{x^4+y-11x^2y^2+y^4}}} , I see some squares, but I do not immediately see what to do with them.


2. {{{x^4+y^2+25}}} has three squares: {{{x^4=(x^2)^2}}} , {{{y^2=(y)^2}}} , and {{{25=5^2}}}
 
3. {{{x^4+y-11x^2y^2+y^4}}} has the squares {{{x^4=(x^2)^2}}} and{{{y^4=(y^2)^2}}} .