Okay I have 2 problems I have worked on and have stumped me. I may throw a bit of fg your way if you help me understand.

Here is my work so far:
#1:
So I took the derivative of the first expression (labeled M) wrt (with respect to) y and the second expression (labeled N) wrt x and they equal, thus the equation is exact.
Now I integrate the first expression wrt x and tack on h(y). Then I derive this expression wrt y.
At this point I should be able to set the expression I've gotten equal to the original N expression and cancel some stuff out to solve for h'(y), but I can't because there are x's left over I can't deal with
Answer in book: [x-(24x^3+x^2-8x-16)^1/2]/4



#2:
If you can't read the problem it says (e^x+1)dy/dx = y-ye^x
I rearranged things to be in the standard form of a separable ODE.
Integrate both sides, right side was hard to integrate, but I checked it with an integral calculator online.
Solve for y by taking e to the power of everything gives me the last line.
What I have is not even close to what the book has....
Answer in book: y = c/cosh^2(x/2) <- Where the fuck did h even come from?

I'm not really helping but for #2 I'm pretty sure it's hyperbolic cos and not a new variable :

http://www.math10.com/en/algebra/hyperbolic-functions/hyperbolic-functions.html

I would write :
(9x² + y - 1).dx - (4y - x).dy = 0, and y(1) = 0
(9x² - 1).dx - 4y.dy + ( y.dx + x.dy ) = 0
d(3x³ - x) - d(2y²) + d(xy) = 0
d(3x³ - x + xy) = d(2y²)
Hence,
3x³ - x + xy = 2y² + C
where C is some constant.

Solve this for y (quadratic equation) :

2y² - xy - 3x³ + x + C = 0

Δ = (- x)² - 4(2.(- 3x³ + x + C))
Δ = 24x³ + x² - 8x - 8C

y(x) = [ x ± √(24x³ + x² - 8x - 8C) ] / 4

Now express the fact that y(1) = 0 :

[ 1 ± √(17 - 8C) ] / 4 = 0

check that C = 2, and that the sign "minus" is the only possibility :

y(x) = [ x - √(24x³ + x² - 8x - 16) ] / 4

This should match the answer in your book.

#2 :

Starting from :

ln y = x - 2.ln (1 + exp x ) + C

Notice that 1 + exp x > 0 for all x, so the absolute value can be taken out.

y = exp ( ln y ) = exp ( x - 2.ln (1 + exp x ) + C )
y = exp x . exp ( - 2.ln (1 + exp x )) . exp C
y = exp x . exp C . ( exp ( ln ( 1 + exp x ) ) ) ^ (-2)
y = exp x . exp C . ( 1 + exp x ) ^ (-2)

Since C is a random constant, so is exp C ( a random positive constant), so I'll denote also C what was previoulsy called exp C :

y = C. exp x / ( 1 + exp x )²

This is your answer.
Now you can use a different function to express it :

cosh u = ( exp u + exp (-u) ) / 2
for all real u.
This function is the hyperbolic cosine ("h" stands for "hyperbolic").

https://en.wikipedia.org/wiki/Hyperbolic_function

To transform the expression of y, try to factorize by the exponential of the mean :
1 + exp x = exp 0 + exp x = exp (x/2) . ( exp (-x/2) + exp (x/2) )
x/2 being the mean between 0 and x.

It appears that exp (-x/2) + exp (x/2) is the double of cosh x , and finally :
1 + exp x = 2 . exp (x/2) . cosh (x/2)

Replace in the previous expression of y :

y = C . exp x / ( 2 . exp (x/2) . cosh (x/2) )²
y = (C/4) / ( cosh (x/2) )²

and again, C/4 is a random positive constant, that you can as well call C :

y = C / ( cosh (x/2) )²

Okay, thank you I understand this one. So I was there on #2 other than a slight mistake, I have just never been taught anything about hyperbolic functions, weird that they would assign this problem...

However I am still confused on #1, I don't understand the steps you are doing at the beginning. The first step you do gives (9x² - 1).dx - 4y.dy + ( y.dx + x.dy ) = 0

I don't know how you rearranged the problem to arrive at this and the next couple steps.

You made a mistake in your partial derivative wrt to y on #1
should be end up like this

x+h'(y)=-4y+x

h'(y)= -4y
h(y)=-2y^2

Shit That's what happens when I do math at 4am.

Solved them both!

Donated fg to each of you