Problem:
Let $f:[0,1] \rightarrow \Bbb R$ be a continous function that is differentiable on $(0,1)$, and where $f(0)=1$ and $f(1)=e$. Show that there exists $\zeta\in(0,1)$ such that $f(\zeta)=f'(\zeta)$.
So I did this problem with my professor. I first said that I would approach this problem by proving that the only function that fulfills the said function, obviously, is $f(x)=e^x$. (I happened to have seen the proof of that here on MSE). Then he said 'No, don't do that, use something you have learned during the lectures'.
So all we learned in Differentiation was the Mean Value Theorem, Intermediate Value Theorem, and Rolle's Theorem. So in my head, it was obvious that I needed to come up with the solution with one or more of these theorems. I thought for a while, couldn't come up with an answer, and then he gave me a hint: he said you can use Rolle's Theorem to solve it. Then again, I thought for a while and could not think of anything.
Then he gave me another hint: he said to define another function so that you can use Rolle's Theorem. 'You already know the function is $f=e^x$, the new function combined with Rolle's Theorem should help you solve it.' I still don't know how (probably got nervous and couldn't think anymore). Then he gave me the new function:
$$g(x):=f(x)e^{-x}$$ Then I was able to solve the problem by using Rolle's Theorem. It was pretty clear then.
So my question is: How do you go about this kind of problem? I have never seen this type of problem before. I am now afraid that even if I do more and more problems, I am only able to solve similar types of problems, and that if I see something new, I would be lost again. How can I improve?
And I want to do more of these kind of problems, but unfortunately I cannot find many of these kinds of problems.
I know this might not be the type of question that I should ask on MSE, but I really don't know what else to do.
Solving the specific problem itself
First notice that $f(x)=e^x$ is not the only function satisfying these conditions. There are an infinite number of smooth curves joining the points $(0,1)$ and $(1,e)$ which are the graph of a function satisfying the conditions in the question.
For a motivated solution to the problem, since we wish to show that there exists $\zeta$ such that $f(\zeta)=f'(\zeta)$ it makes sense to move all expressions to one side of the equation in such a way that Rolle's theorem would help us. Here we get $\frac{f'(\zeta)}{f(\zeta)}=1$ and then $\frac{f'(\zeta)}{f(\zeta)}-1=0$. (We do not have to worry about division by the possibly vanishing $f(\zeta)$ here-- this is just to motivate the definition of a function which we will use later on, and there will be no division by $0$ in the solution.)
To use Rolle's theorem we need $\frac{f'(\zeta)}{f(\zeta)}-1$ to be the derivative of some function, and we recognize the logarithmic derivative on the left so it makes sense to define $g(x)=\log(f(x))-x$. There is a little problem though-- in order to use Rolle's theorem we need $g(x)$ continuous on $[0,1]$ and it is possible that $f(x)\leq 0$ for some $x\in[0,1]$, which would make our $g(x)$ undefined and certainly not continuous. To fix the problem with the domain of $\log x$, define instead $h(x)=e^{g(x)}$ so that we get $h(x)=f(x)e^{-x}$ which is precisely the function your professor advised you to use.
Now $h(0)=1$, $h(1)=1$, and $h(x)$ is continuous on $[0,1]$ and differentiable on $(0,1)$ so by Rolle's theorem there exists $\zeta\in(0,1)$ such that $h'(\zeta)=0$. This means $f'(\zeta)e^{-\zeta} - f(\zeta)e^{-\zeta} = 0$ so $f'(\zeta)=f(\zeta)$ as required.
General remarks
There is a more general question in your post which is how to approach these type of questions in general. It is not possible to give a complete recipe and it is hard to pinpoint what can be meant by 'these type' of questions. However, when it seems like your problem requires the usage of some theorem which needs to operate on a function (such as the intermediate value theorem, Rolle's theorem, or the mean value theorems), it can often be useful to define a function by taking the expression in the statement of that which you are trying to prove and moving everything to one side to get $h(x)=0$, so that the appropriate theorem can be used for this $h(x)$. In applications of Rolle's theorem there is the extra complication that you want the expression on the left hand side to be the derivative of some function, so you define your function to be the antiderivative of the left hand side, as was done above.
If the function you defined does not satisfy the required conditions to use the appropriate theorem then you need to either find another way to move all expressions to one side of the equation (e.g. with subtraction instead of division, or the other way around), or else apply some other trick like the one above where we used exponentiation to get rid of the problem with continuity.