On
$f(f^{-1}(x))=x$
The right hand side is always increasing in x so the left hand side must also be increasing in x. You know that $f(\cdot)$ is increasing. What does that mean for $f^{-1}(x)$?
On
Let $f\colon A\to B$ an increasing and bijective function. Suppose $f^{-1}$ is not increasing. There exists $y_1$ and $y_2$ in $B$ such that $y_1 < y_2$ and $f^{-1}(y_1) \geqslant f^{-1}(y_2)$. As $f$ is increasing, we have $f(f^{-1}(y_1)) \geqslant f(f^{-1}(y_2))$, that is $y_1\geqslant y_2$. Contradiction: the function $f^{-1}$ is increasing.
On
That $f$ is increasing means that $x\leq y\to f(x)\leq f(y)$ holds. Then also $x<y\to f(x)<f(y)$ since $f$ is injective, as well as $f(y)<f(x)\to y<x$ by contrapositive, which is the same as $f(x)<f(y)\to x<y$ (only the names of the variables were interchanged). So $f$ increasing means $x<y\leftrightarrow f(x)<f(y)$.
Now if $f$ has inverse $f^{-1}$, one has $$ \hbox{$f$ increasing}\iff x<y\leftrightarrow f(x)<f(y) \implies f^{-1}(a)<f^{-1}(b)\leftrightarrow f(f^{-1}(a))<f(f^{-1}(b)) \\ \iff f^{-1}(a)<f^{-1}(b)\leftrightarrow a<b \iff \hbox{$f^{-1}$ increasing}, $$ where the implication step specialises $x=f^{-1}(a)$ and $y=f^{-1}(b)$.
On
Suppose by way of contradiction that $f^{-1}$ were not increasing.
That is, suppose that there were $y_1<y_2$ in the range of $f$ for which $f^{-1}(y_1)\ge f^{-1}(y_2)$. Applying $f$, we get $f(f^{-1}(y_1))\ge f(f^{-1}(y_2))$. Cancelling out $f$ and $f^{-1}$ we get $y_1 \ge y_2$, a contradiction to our assumption that $y_1<y_2$.
So $f^{-1}$ is increasing.
A non-derivative proof.
What we have is $\forall u,v. (u>v\implies f(u)>f(v))$.
We want to prove $\forall u,v. (f(u)>f(v)\implies u>v)$.
Suppose the opposite, that there exist $u_0,v_0$ such that $f(u_0)>f(v_0)$ is true and $u_0>v_0$ is false for the sake of contradiction.
Since $u_0>v_0$ is false, $u_0=v_0$ or $u_0<v_0$.
Case 1. $u_0=v_0$. Then $f(u_0)=f(v_0)$, contradiction.
Case 2. $u_0 < v_0$. Then since $\forall v,u. (v>u\implies f(v)>f(u))$, we have $f(v_0)>f(u_0)$, contradiction.
Hence there cannot exist such $u_0,v_0$.