Solving functional equations - learn by examples

Q1.
Find all f: R -> R s.t. for all real x,y 
f(x + y) + f(xy) = x + y + xy + 2
S1:
plug in y = 0
=> f(x) + f(0) = x + 2.
Plugin x = 0
=> f(0) = 1
=> f(x) = x + 1

Check the answer in original equation.
f(x+y) = x + y + 1
f(xy) = xy + 1
They satisfy the original equation.

Q2.
Find all f: R -> R s.t. for all real x,y
f(2x + y) = f(-y) + 2f(x) + 4y
S2.
We want to find f(x) but other two f() are in our way. Remove them.
2x + y = -y => -y = x so use this substitution y = -x
f(x) = f(x) + 2f(x) + 4y
=> f(x) = -2y = 2x
Check answer in the original equation. It matches.

Q3.
Find all f: R -> R s.t. for all real x,y
f(f(x) + y) = f(x-y) + 2y
S3.
Making inputs inside f() equal
=> f(x) + y = x - y 
=> 2y = x - f(x)
=> f(x) = x
And this matches in the original equation.

Q4.
Find all f: R -> R s.t. for all real x,y
2f(x) + f(1-x) = 4x + 1
S4.
Replace x with 1-z
2f(1-z) + f(z) = 4(1-z) + 1
Now we have 2 vars f(z), f(1-z) and 2 equations. Solve to get f(x).
f(x) = 4x - 1 = answer

Q5.
Find all f: R -> R s.t. for all real x,y
f(x) + xf(y) = x + f(x)f(y)
S5.
Swap x,y and solve resulting 2 equations to get:
(1-y)f(x) + (x-1)f(y) = x - y
Put x = 1 and y != 1 you will get f(1) = 1
In original equation put y = 1 and you will eventually get f(x) = 1 or f(x) = x as answers.

Q6.
Find all f: R -> R s.t. for all real x,y
f(x+y) + xy = f(x)f(y)
S6.
Put x = y = 0 to get f(0) = 0 or f(0) = 1.
Now put f(0) = 0 and y = 0 in the original equation you will get:
f(x) = 0
But putting this in the original equation doesnt work.
Then start with f(0) = 1 and eventually you get:
f(x) = 1-x and f(x) = x + 1