Showing monotonicity of sinh, cosh and tanh 
a) Show that $x>0\implies \sinh(x)>0$

$\sinh(x)=\frac{e^x-e^{-x}}{2}$
Let $x>0$.
$\frac{e^x-e^{-x}}{2}>0$
$\implies e^x-e^{-x}>0$
$\implies e^x>e^{-x}$
$\implies x>-x$ since taking the natural logarithm doesn't change the inequality sign.
$\implies 2x>0$
$\implies x>0$
$\implies 1>0$
Which is true

b) Conclude that $\cosh$ on $\mathbb{R}^{+}$ and $\sinh, \tanh:=\frac{\sinh}{\cosh}$ on $\mathbb{R}$ are strictly monotone increasing.

$\cosh(x)=\frac{e^x+e^{-x}}{2}$
Let $0<x<y$.
$\frac{e^x+e^{-x}}{2}<\frac{e^y+e^{-y}}{2}$
$\implies e^x+e^{-x}<e^y+e^{-y}$
$\implies e^x-e^{-y}<e^y-e^{-x}$
$\implies e^x(1-e^{-x-y})<e^y(1-e^{-x-y})$
Since the stuff in the brackets is the same we only have to show
$e^x<e^y$ where we take the natural logarithm and get
$x<y$
Which is true. Therefore $\cosh$ is strictly monotonic increasing.
Let $x,y\in\mathbb{R}$ and $x<y$ 
$\sinh(x)=\frac{e^x-e^{-x}}{2}$
$\frac{e^x-e^{-x}}{2}<\frac{e^y-e^{-y}}{2}$
$\implies e^x-e^{-x}<e^y-e^{-y}$
$\implies e^x+e^{-y}<e^y+e^{-x}$
$\implies e^x(1+e^{-y-x})<e^y(1+e^{-y-x})$
Which is analogous to $\cosh$. Therefore $\sinh$ is strictly monotonic increasing.
$\tanh(x)=\frac{\sinh(x)}{\cosh(x)}=\frac{\frac{e^x-e^{-x}}{2}}{\frac{e^x+e^{-x}}{2}} <\frac{\frac{e^y-e^{-y}}{2}}{\frac{e^y+e^{-y}}{2}}=\tanh(y)$
$\implies \frac{e^x-e^{-x}}{e^x+e^{-x}}<\frac{e^y-e^{-y}}{e^y+e^{-y}}$
can we use at the end maybe a WLOG-argument, by setting $x=0$, and $y>0$?
Also is the above correct so far?
 A: All your proofs are logically incorrect because they are backwards. (It is a very common logical mistake among students.) The statement that you need to show must be the result of your proof, not its starting point. It looks to me that you do have all the correct ingredients, you just need to fix the logic. Two possible ways to do that:


*

*demonstrate that each implication is in fact an equivalence (if that's the case);

*set up a correct proof by rewriting your work backward, i.e. start with a true statement (such as "$1>0$", for example) and show that the desired statement can be deduced from it.
Note 1: there are may be some minor arithmetic/calculus omissions, but you can take care of those after you fix the logic.
Note 2: using calculus and derivatives can help with some parts of your question, unless you're not allowed to use calculus.
A: One answer has already pointed out the main structural error in your proofs
(implication in the wrong direction).
Another thing to watch out for is the direction of inequalities when
multiplying both sides by some quantity.
In one of your steps you write

$\implies e^x(1-e^{-x-y})<e^y(1-e^{-x-y})$
Since the stuff in the brackets is the same we only have to show
$e^x<e^y$

Aside from the fact that your proof actually should show $e^x < e^y$ first
and then conclude that $e^x(1-e^{-x-y})<e^y(1-e^{-x-y})$,
that conclusion is only true if $(1-e^{-x-y}) > 0$.
In particular:


*

*If $(1-e^{-x-y}) = 0$ then $e^x(1-e^{-x-y})=e^y(1-e^{-x-y})$.

*If $(1-e^{-x-y}) < 0$ and $e^x < e^y$  then $e^x(1-e^{-x-y}) > e^y(1-e^{-x-y})$.


It should be easy enough to show that $(1-e^{-x-y}) > 0$
since it is already given that $0 < x < y$,
and at some level of math we might not even expect to see it mentioned,
but if you're just learning to write proofs you should make sure
that whoever reads the proof can see that you didn't skip this step.
(Besides, if we just reverse the implication arrows in your proof
we might think we can remove the condition $0<x$ and
prove $\cosh x$ is increasing even for negative $x$, which is not true.)

Another thought: the word "conclude" in part (b) suggests that you
were intended to use the fact that $\sinh x$ is increasing to
prove that $\cosh x$ is increasing,
perhaps by appealing to the identity $\cosh^2 x - \sinh ^2 = 1,$
rather than going through the
whole exercise with $e^x$ and $e^{-x}$ instead.
Honestly, it's not clear to me how that's easier or simpler, but
it does seem to be the idea.
For $\tanh x$ increasing, I think the use of the identity
$\tanh^2 x + \mathrm{sech}^2\, x = 1$ (where $\mathrm{sech}\,x = 1/\cosh x$)
might actually be easier than fiddling with the powers of $e$.
