I'm about to learn about differentiating inverse functions at school, and the formula we're being told we'll be using is [assuming $g(x)=f^{-1}(x)$]:
$$f'(x)=\frac{1}{g'(f(x))}$$
In other places online, however, I have seen a much simpler formula, $dx/dy * dy/dx = 1$. (I am quite interested in Leibniz notation, as the old intuitive ideas often stun me in their elegance.)
However, here I cannot for the life of me figure out the connection between the Leibniz notation above and the more complicated "prime" notation above. I have searched far and wide across the Internet, but am still confused. Can anyone explain? How would I calculate the derivative of an inverse function using Leibniz notation, and how does that connect to the process of computing an inverse function's derivative using "prime" notation?
I think that your teacher needs to clarify the notation with examples.
Definition: Let $g(x)$ be injective. Then $f(x)$ is an inverse function of $g(x)$ provided
$$g(f(x)) = f(g(x)) = x$$
Example: Suppose $g(x)=2x-1$, which is injective. Then, $g^{-1}(x)$ can be found by switching $x$ and $y$ and then solving for $y$
$$x=2y-1\implies y=\frac{x+1}{2} \implies g^{-1}(x)=\frac{x+1}{2}$$
Alternatively, if we define $f(x)=\dfrac{x+1}{2}$ then
$$g(f(x))=g\left(\frac{x+1}{2}\right)=2\left(\frac{x+1}{2}\right)-1=x$$
and
$$f(g(x))=f\left(\frac{(2x-1)+1}{2}\right)=x$$
so we see that the definition holds in this example. This leads to the result provided by your course notes.
Theorem: Let $g(x)$ be a function that is differentiable on an interval $I$. If $g(x)$ has an inverse function $f(x)$, then $f(x)$ is differentiable at any $x$ for which $g'(f(x))\neq 0$. Moreover,
$$f'(x)=\frac{1}{g'(f(x))},~~g'(f(x))\neq 0$$
To prove this theorem, we start with the definition
$$g(f(x)) = x$$
and then differentiate implicitly
$$\frac{d}{dx}\Big[g(f(x))\Big]=\frac{d}{dx}(x)$$
where we set $y=g(u)$ and $u=f(x)$ and then use the chain rule
$$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=g'(u)f'(x)=g'(f(x))f'(x)$$
and since $\frac{d}{dx}(x)=1$, we know that
$$g'(f(x))f'(x)=1$$
then since $g'(f(x))\neq 0$, we can divide by it to form
$$f'(x)=\frac{1}{g'(f(x))}$$
Example: Suppose $x>0$ and
$$g(x)=x^2$$
then $f(x)=\sqrt{x}$ is its inverse. We have
$$f'(x)=\frac{1}{2\sqrt{x}}$$
and
$$g'(x)=2x$$
therefore
$$g'(f(x))=2(f(x))=2\sqrt{x} \implies \frac{1}{g'(f(x))}=\frac{1}{2\sqrt{x}}$$
so we see that
$$f'(x)=\frac{1}{g'(f(x))}$$
In terms of Leibniz notation, we can adjust the proof so that all of the primes are replaced by differentials. In this notation $$f'(x)=\frac{df(x)}{dx}=\frac{df}{dx}$$
and letting $y=f(x)$ and $x=g(y)$ forms
$$\frac{dy}{dx}=\frac{df}{dx},~~g'(f(x))=\frac{dg(y)}{dy}=\frac{dx}{dy}$$
so that
$$\frac{dy}{dx}\frac{dx}{dy}=1 \implies \frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$$
or
$$\frac{df}{dx}=\frac{1}{\frac{dg(y)}{dy}}$$
and we once again arrive at
$$f'(x)=\frac{1}{g'(f(x))}$$
as desired.
Example: Suppose that $y=f(x)=e^x$. Then, we want to show that
$$f'(x)=\frac{1}{g'(f(x))}$$
We are given
$$\frac{df}{dx}=\frac{dy}{dx}=e^x$$
Next, we can deduce that $x=g(y)=\ln(y)$ so that $g$ is the natural log function (the inverse of $e^x$ is the natural log function). Then
$$g'(f(x))=\frac{dx}{dy}=\frac{dg(y)}{dy}=\frac{d}{dy}\ln(y)=\frac{1}{y}=\frac{1}{e^x}$$
Therefore,
$$\frac{1}{g'(f(x))}=e^x$$
In terms of prime notation versus Leibniz notation, it is largely a matter of personal preference. Many people prefer using the Leibniz notation because the chain rule can be easily identified as
$$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$$
instead of
$$\Big[f(g(x))\Big]' = f'(g(x))g'(x)$$
There is also the additional advantage that writing
$$\frac{dy}{dx}$$
makes it abundantly clear that $x$ is the independent variable and $y$ is the dependent variable. This can sometimes be confusing when you have multiple variables with primes.
I often switch between the two notations as needed. It might be best to work with both notations if the primes confuse you.